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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
0.45 g of an organic compound gave 0.44 g of `CO_(2)` and 0.09 g of `H_(2)O`. The molecular mass of the compound is 90 amu. Calculate the molecular formula. |
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Answer» Correct Answer - `C_(2)H_(2)O_(4)` Step I. Percentage of elements Percentage of carbon `= (12)/(44) xx ("Mass of "CO_(2))/("Mass of compound")xx100` `=(12)/(44)xx(0.44)/(0.45)xx100=26.67` Percentage of hydrogen `=(2)/(18)xx("Mass of" H_(2)O)/("Mass of compound")xx100` `=(2)/(18)xx(0.09)/(0.45)xx100=2.22` Percentage of oxygeb `= 100 - (26.67 + 2.22) = 100-28.89 = 71.11` Step II. Empirical formula of the compound `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",26.67,12,(26.67)/(12)=2.22,(2.22)/(2.22)=1.0,1),("H",2.22,1,(2.22)/(1)=2.22,(2.22)/(2.22)=1.0,1),("O",71.11,16,(71.11)/(16)=4.44,(4.44)/(2.22)=2.0,2):}` Empirical formula of the compound `= CHO_(2)` Step III. Molecular formula of the compound Empirical formula mass `= 12+1 +2 xx 16 = 45 u` Molecular mass = 90 amu (Given) `n=("Molecular mass")/("Empirical formula mass")=((90u))/((45u))=2` Molecular formula `= n xx` Empirical formula `= 2 xx CHO_(2)=C_(2)H_(2)O_(4)`. |
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| 202. |
An organic compound on analysis gave the following percentage composition : `C = 57.8 %, H= 3.6 %` and the rest is oxygen. The vapour density of the compound was found to be 83. Find the molecular formula of the compound. |
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Answer» Correct Answer - `C_(8)H_(6)O_(4)` Step I. Percentage of oxygen `100 - (57.8 + 3.6) = 100 - 61.4 = 38.6` Step II. Empirical formula of organic compound `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",57.8,12,(57.8)/(1)=4.82,(4.82)/(2.41)=2.0,4),("H",3.6,1,(3.6)/(1)=3.6,(3.60)/(2.41)=1.5,3),("O",38.6,16,(38.6)/(16)=2.41,(2.41)/(2.41)=1.0,2):}` Empirical formula of the compound `= C_(4)H_(3)O_(2)` Step II. Molecular formula of the compound Empirical formula mass `= 4 xx 12 + 3xx 1+ 2 xx 16 = 83 u` Molecular mass `= 2 xx V.D = 2 xx 83 = 166u` `n=("Molecular mass")/("Empirical formula mass")=((166u))/((83u))=2` `:.` Molecular formula of compound `= 2 xx C_(4)H_(3)O_(2) = C_(8)H_(6)O_(4)`. |
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| 203. |
Fill in the blanks. a. `2.24 L` ammonia at `STP` neutralised `100 mL` of a solution of `H_(2) SO_(4)`. The molarity of acid is…….. b. The equivalent weight of a metal carbonate `0.84 g` of which reacts exactly with `40 mL` of `N//2 H_(2) SO_(4)` is ........ c. `1.575 g`, of hydrated oxalic acid `(COOH)_(2). nH_(2) O` is dissolved in water and the solution is made to `250 mL` On titration, `16.68 mL` of this solution is required for neutralisation of `25 mL` of `N//15 NaOH`. The value of water crystallisation, i.e., `n`, is............ d. `1 mL` of `H_(3) PO_(4)` was diluted to `250 mL`. `25 mL` this solution requried `40.0 mL` of `0.10 N NaOH` for neutralisation using phenolphthanlen as indicator. The specific gravity of acid is.............. The density of 1.48 mass percent calcium hydroxide solution is `1.25 g mL^(-1)`. The volume of `0.1 M HCl` solution required to neutralise `25 mL` of this solution is........ |
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Answer» Correct Answer - A::B::C::D a. Eq of `NH_(3) =` `("Volume of a gas at" STP)/("Volume of 1 equivalent of gas") = (2.24)/(22.4) = 0.1 "equivalent"` mEq of `NH_(3) = 0.1 xx 10^(3) = 100` mEq of `H_(2)SO_(4) = 100 xx N` mEq of `H_(2) SO_(4) = mEq "of" NH_(3)` `100 xx N -= 100` `N_(H_(2)SO_(4)) = 1` `M_(H_(2)SO_(4)) = (1)/(2) = 0.5 M` b. mEq of `MCO_(3) -= mEq "of" H_(2)SO_(4)` `[0.84)/(Ew(M) + Ew (CO_(3)^(2-))] xx 10^(3) = 40 xx (1)/(2)` `((0.84)/(E + (60)/(2))) xx 10^(3) = 20` `((0.84)/(E + 30)) = (20)/(1000)` `E = 12` `Ew` of metal carbonate `= 12 + 30 = 42` `Mw` of `(COOH)_(2), nH_(2) O = 90 + 18 n` `[{:("Eq of" (COOH)_(2).nH_(2)O),("in" 250 mL "solution"):}] = ((1.575)/(90 + 18 n)) xx 2` (`n` factor = 2) .....(i) `16.68 mL xx N = 25 xx (1)/(15)` `N` (acid) `= 0.099 ~~ 0.1 Eq L^(-1)` `(0.1 xx 250)/(1000)` `= (0.1)/(4)` Eq per `250 mL` Equating equation (i) and (ii), we get `((2 xx 1.575)/(90 + 18 n)) = (0.1)/(4)` Solve for `n`, ` n = 2` Formula : `(COOH)_(2) . 2H_(2) O` d. mEq `H_(3)PO_(4) mEq "of" NaOH` `25 mL xx N = 40 mL xx 0.1 N` `N "of" H_(3) PO_(4) = 0.16` `N = (W_(2) xx 1000)/(Ew_(2) xx V_(sol)) (Mw H_(3) PO_(4) = 98, Ew = (98)/(3))` (`n` factor = 3) `d_(H_(3)PO_(4)) = (W_(2) xx 1000)/(98//3 xx 250), Wt H_(3)PO_(4) mL^(-1) = 1.32 g mL^(-1)` e. `N_(Ca(OH)_(2)) = (% "by Weight"xx 10 xx d)/(Ew_(2))` `= (1.48 xx 10 xx 1.025)/(37)` mEq of `Ca (OH)_(2) -= mEq "of" HCl` `25 mL xx 0.41 N -= 0.1 xx V_(HCl)` `V_(HCl) = 10.25 mL` |
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| 204. |
Ammonia in `0.224 g` of a compound `Zn(NH_(3))_(x)Cl_(2)` is neutralised by `30.7 mL` of `0.20 M HCl`. The value of `x` in the formula isA. 4B. 5C. 6D. 8 |
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Answer» Correct Answer - C `Mw "of" Zn (NH_(3))_(x) Cl_(2) = 65.30 + 17x + 35.4 xx 2` `= 136.30 + 17 x` `(17 x + 36.30)g` of compound contains `= x` mole of `NH_(3)` `0.224 g` compound `= (X)/(17 + 136.30) xx 0.224` `x` mol of `NH_(3) = x eq "of"NH_(3)` Eq of `NH_(3) = Eq "of" HCl` `(0.244 x)/(17x + 136.30) = (30.7 xx 0.2)/(1000) eq "of"HCl` `x ~~ 6.75 ~~ 6` (as the choice given) |
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| 205. |
`0.44 g` of a hydrocarbon on complete combustion with oxygen gave `1.8 g` of water and `0.88 g` of carbon dioxide. Show that these result are in agreement with the law of conservation of mass. |
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Answer» Mass of carbon in `0.88 g` of `CO_(2)=(12)/(44) xx 0.88 g = 0.24 g` Mass of hydrogen in `1.8 g` of `H_(2)O=(2)/(18)xx1.8 g = 0.20 g` Mass of carbon and hydrogen present in the products `= 0.24+0.20 = 0.44 g` Mass of hydrocarbon `(C_(x)H_(y))` reacted `= 0.44 g` Since there is no change in the total mass of carbon and hydrogen as a result of the reaction, this proves the law of conservation of mass. |
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| 206. |
A compound on analysis gave the following percentage composition by weight: hydrogen = 9.09, oxygen = 36.36 carbon = 54.55 Its `VD` is 44. Find the molecular formula of the compound. |
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Answer» Correct Answer - B::C::D `{:("Element", "Moles", "Least ratio"),(C, (54.55)/(12) = 4.54, (4.54)/(2.27) = 2),(H, (9.09)/(1) = 9.09, (9.09)/(0.27) = 4),(O, (36.36)/(16) = 2.27, (2.27)/(2.27) = 1):}` Empirical formula `= C_(2) H_(4) O`. Empirical formula weight `= 12 xx 2 + 1 xx 4 + 16 = 44` `VD = 44`, Molecular Weight `= 2 xx VD = 2 xx 44 = 88` `n = ("Molecular Weight")/("Empirical formula weight") = (88)/(44) = 2` `= 2 xx (C_(2) H_(4) O) = C_(4) H_(8) O_(2)` |
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| 207. |
Two litre of `NH_(3)` at `30^(@)C` and 0.20 atmosphere is neutralised by `134 mL` of a solution of `H_(2) SO_(4)`. Calculate the normality of `H_(2) SO_(4)`. |
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Answer» Correct Answer - A::B `PV = (W)/(Mw) RT` Mole `= (W)/(Mw) = (PV)/(RT) = (0.2 xx 2)/(0.0821 xx 303) = 0.01608 mol` = 16.08 mmol or mEq `({:(1 "mole of" NH_(3) = 1 g Eq. of NH_(3)),(1 "mmole of" NH_(3) = 1 mEq of NH_(3)):})` `N xx V = mEq` `N xx 134 mL = 16.08` `:. N = 0.12` |
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| 208. |
Five grams of `KCIO_(3)` yield `3.041 g` of `KCI` and `1.36 L` of oxygen at standard temperature and pressure. Show that these figures support the law of conservation of mass within limits of `+- 0.4 %` error. |
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Answer» According to gram-molecular volume law, `22.4 L` of all gases and vapours at `STP` weigh equal to their molecular weights denoted in grams. `:.` Weight of `1.36 L` of oxygen at `STP = (32 xx 1.36)/(22.4) = 1.943 g` Weight of `KCl` formed `= 3.041 g` (given) `:.` Total weight of product `(KCl + O_(2))` `= 3.041 + 1.943 = 4.984 g` Error `= 5 - 4.984 = 0.01 g` `%` error `= (0.016 xx 100)/(5) = 0.32` Hence, the law of conservation of mass is valid within limits of `- 0.4%` error. Thus, the law is supported. |
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| 209. |
`0.22 g` of a hydrogen (i.e., a compound conatining carbon and hydrogen only) on complete combustion with oxygen gave `0.9 g` water and `0.44 g` carbon dioxide. Show that these results are in accordance with the law of conservation of mass (atomic mass of `C = 12, H = 1, O = 16)`. |
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Answer» Carbon and hydrogen in the hydrocarbon on combustion form carbon dioxide and water. From the formula of `CO_(2)` and `H_(2) O`, the weights of carbon and hydrogen contained in `0.44 g CO_(2)` and `0.9 g` water, respectively, can be calculated as under: ltbr Molecular weight of `CO_(2) = 12 + 32 = 44` Molecular weight of `H_(2)O = 2 + 16 = 18` Weight of carbon in `0.44 g` of `CO_(2) = (12 xx 0.44)/(44) = 0.12 g` Weight of hydrogen in `0.9 g` of `H_(2) O = 2 xx (0.9)/(18) = 0.10 g` Total weight of `C` and `H` in the hydrocarbon after combustion `= 0.12 + 0.10 = 0.22 g` Since the weight of carbon hydrogen after combustion is the same as the weight of hydrocarbon (containing carbon and hydrogen only) after combustion, the results are in accordance with the law of conservation of mass. |
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| 210. |
`49 g` of `H_(2) SO_(4)` is disslved in enough water to make one litre of a soltuion of density `1.049 g c c^(-1)`. Find the molarity, normality, moality, and mole fraction of `H_(2) SO_(4)` in the solution. |
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Answer» Correct Answer - A i. `M = (W_(2) xx 100)/(Mw_(2) xx V_(sol) ("in" mL)) = (49 xx 1000)/(98 xx 1000) = 0.5 M` ii. `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) ({:(W_(1)=V_("Sol")xxd_("sol")-W_(2)),(=1000xx1.049-49),(=1000g):})` `= (49 xx 1000)/(98 xx 1000) = 0.5 m` iv. `X_(2) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2))) = (49 // 98)/((100)/(18) + (49)/(98)) = (0.5)/(55.5 + 0.5)` `(0.5)/(56) = 0.009` |
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| 211. |
If `2.4gm` Mg is treated with `0.64 gm O_(2)` Find composition of final product mixture ? [Atomic Mass `Mg = 24]` . |
| Answer» `{:(,Mg(s),+,(1)/(2)O_(2)(s),rarr,MgO(s)),(t=0,0.1mol,,0.2mol,,0),([O_(2)isLR],,,,,),(t=oo,0.1-0.02xx2,,0,,0.02xx2),(,=0.06mol,,=0.04mol,,):}` . | |
| 212. |
Calculate (a) the actual volume of a molecule of water (b) the radius of a water molecule assuming to be spherical (density of water `= 1 g cm"^(-3)`) |
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Answer» (a) Density of water `= 1g cm^(-3)` Gram molecular mass of water `= 18 g` Gram molecular volume `= ((18g))/((1g cm^(-3)))=18 cm^(3)` `6.022 xx 10^(23)` molecules of water occupy volume `= 18 cm^(3) = 18 c c` 1 molecule of water occupies volume `= (18)/(6.022xx10^(23))=2.99xx10^(-23) c c` (b) Volume of a sphere `= 4//3 pi r^(3) :. 4//3 pi r^(3) = 2.99 xx 10^(-23) c c` or `r^(3)=(2.99xx10^(-23)xx3)/(4xx3.143)" "[becausepi=22//7=3.143]` or `r=((2.99xx10^(-23)xx3)/(4xx3.143))^(1//3)=(7.13xx10^(-24))^(1//3)=1.925xx10^(-8)"cm"=1.925 Ã…`. |
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| 213. |
The cost of table salt `(NaCl)` and table sugar `(C_(12)H_(22)O_(11))` are Rs 2 per kg and Rs. 6 per kg respectively. Calculate their cost per mole. |
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Answer» (a) Cost of table salt `(NaCl)` per mole Gram molecular mass of `NaCl=23+35.5 = 58.5 g` Now, 1000 g of NaCl cost = Rs. 2 `:. 58.5 g` of NaCl will cost `= (2)/((1000 g))xx(58.5 g)=0.117` Rupee `=0.117 xx 100 = 12` paise (approx) (b) Cost of table sugar `(C_(12)H_(22)O_(11))` per mole Gram molecular mass of `C_(12)H_(22)O_(11)=12xx12+22xx1+16xx11=144+22+176=342`g Now, 1000g of sugar cost = Rs. 6 `:. 342g` of sugar will cost `= (6)/((1000 g))xx(342 g)=2.052 = 2.0` Rupees (approx) |
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| 214. |
`{:(,2(s),+,3B(g),rarr,4C(g)),(t=0,10mol,,10mol,,):}` Find final moles of `A,B` and `C` . |
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Answer» LR is B `{:(2A,+,3B,rarr,4C),(t=oo10-10((2)/(3)),,0,,10((4)/(3))),(t=oo(1)/(3)mol,,0,((40)/(3))mol,):}` |
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| 215. |
Which of the following reactions occur to completion ? . `{:(,,2A+,3B,rarr,C,+,2D,),((a)"Initial moles",20,,30,,,0,,0),((b)"Initial moles",40,,60,,,10,,0),((c)"Initial moles",40,,90,,,0,,0),((d)"Initial moles",50,,75,,,0,,10):}` |
| Answer» Correct Answer - (a,b,d) | |
| 216. |
Express the following numbers to four significant figures (i) 5.607982 (ii) 32.392800 (iii) `1.78986xx10^(3)` (iv) 0.007837. |
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Answer» (i) `5.607982=5.608` (ii) `32.392800 = 32.39` (iii) `1.78986 xx10^(3) = 1.790 xx 10^(3)` (iv) `0.007837 = 0.007837`. |
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| 217. |
The mass of copper metal is `6.342 g` and the density of copper is `7.6 g//cm^(3)`. What is its volume ? |
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Answer» Mass of copper metal `= 6.342 g` Density of copper metal `= 7.6 g//cm^(3)` Volume `= ("Mass")/("Density")=((6.342g))/((7.6g//cm^(3)))=0.834 cm^(3)` The final result has to be reported upto two significant figures because the least pecise number (7.6) has two significant figures `:.` In the final result, the digit 4 is dropped and the correct answer `=0.83 cm^(3)` |
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| 218. |
Round off the following upto three significant figures (i) `34.246` (ii) `10.4207` (iii) `0.04587` (iv) `10.668`. |
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Answer» (i) `34.2` (ii) `10.4` (iii) `0.0459` (iv) `10.7`. |
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| 219. |
1.61 gm of `Na_(2)SO_(4). 10H_(2)O` contains same number of oxygen atoms as present in:A. 0.98 gm `H_(2)SO_(4)`B. 0.08 gm `SO_(2)`C. 1.78gm `H_(2)S_(2)O_(7)`D. 0.05 gm `CaCO_(3)` |
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Answer» Correct Answer - C |
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| 220. |
Glucose is a physiological sugar. What is the mass% `C` mass% `H` and mass% `O` in glucose `(C_(6) H_(12) O_(6))`? |
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Answer» Correct Answer - A::B::C::D `Mw` of glucose `= C_(6) H_(12) O_(6) = 12 xx 6 + 1 xx 12 + 16 xx 6` `= 180 g` a. `180 g` of glucose `= 72 g of C` `100 g` of glucose `(72 xx 100)/(180) = 40%` b. `% of H = (12 xx 100)/(180) = 6.67 %` c. `% of O = (96 xx 100)/(180) = 53.33 %` |
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| 221. |
The molality of a sulphuric acid solution is `0.2`. Calculate the total weight of the solution having 1000 gm of solvent.A. 1000 gB. `1098.6` gC. `980.4` gD. `1019.6` g |
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Answer» Correct Answer - D |
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| 222. |
The molality of `1 L` solution with `x% H_(2)SO_(4)` is equal to 9. The weight of the solvent present in the solution is `910 g`. The value of `x` is:A. `90`B. `80.3`C. `30.38`D. `46.87` |
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Answer» Correct Answer - D Molality of `H_(2)SO_(4)` is 9 i.e. 9 mole of `H_(2)SO_(4)` in 1kg solvent 1kg solvent contain = 9 mole `H_(2)SO_(4)` 1kg solvent contain `= 9 xx 98 "wt" H_(2)SO_(4)` 1000kg solvent contain `= 9 xx 98//1000 xx 910` 910kg solvent contain `=802.62g` wt. of solution `=910 kg` wt. of solution `802.62+910=1712.62g` x `%` by wt `=("wt of solvent")/("wt of solution")xx100` `=(802.62)/(1712.62) xx 100=46.87` wt of solution `= 802.62 + 910` `=1712.62g` `x%` by wt `x%` ` = ("wt of solute")/("wt of solution") xx 100` `= (802.62)/(1712.62) xx100 = 46.87` . |
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| 223. |
The molality of `1 L` solution with `x% H_(2)SO_(4)` is equal to 9. The weight of the solvent present in the solution is `910 g`. The value of `x` is:A. 90B. 80.3C. 40.13D. 9 |
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Answer» Correct Answer - B Molality `= (1000 xx "Weight of solute")/(Mw xx "Weight of solven")` `9 = (1000 xx W)/(98 xx 910)` `W = 802.6 gL^(-1)` `= 80.26 g` per `100 mL` |
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| 224. |
`4.0 g` of `NaOH` is contained in one decilitre of aqueous solution. Calculate the following in the solution (d of `NaOH` solution `= 1.038 g mL^(-1)`) a. Mole fraction of `NaOH` b. Molartiy of `NaOH` c. Molality of `NaOH` |
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Answer» `M = (W_(2) xx 100)/(Mw_(2) xx V_(sol))` (1 decilitre `= 100 mL`) `= (4 xx 1000)/(40 xx 100) = 1` `m = (W_(2) xx 1000)/(Mw_(2) xx ("Weight of solution - Weight of solute")` `W_(1)` = weight of solvent = Weight of solution - weight of solute `= V_(sol) xx d_(sol) - W_(2)` `= 100 xx 1.038 - 4 = 99.8 g` `:. m = (4 xx 1000)/(40 xx 99.8) = 1.002` `X_(2) = (n_(2))/(n_(1) + n_(2)) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)))` `= (4 // 40)/((99.8)/(18) + (4)/(40)) = (0.1)/(5.54 + 0.1) = 0.0177` |
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| 225. |
Calculate molarity of NaOH in a solution made by mixing 2 L of `1.5 M NaOH,3 L` of 2 M NaOH and 1 L water [Give your answer after multiplying by 10] |
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Answer» Correct Answer - 15 |
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| 226. |
All samples of carbon dioxide contain carbon and oxygen in the mass ratio of `3:8` This is in agreement with the law of .A. conservation of massB. constant proportionC. multiple proportionsD. gaseous volumes |
| Answer» Correct Answer - B | |
| 227. |
The crystalline salt `Na_(2)SO_(4)` on heating loses 55.9 % of its weight. The formula of the crystalline salt is :A. `Na_(2)SO_(4).5H_(2)O`B. `Na_(2)SO_(4).7H_(2)O`C. `Na_(2)SO_(4).2H_(2)O`D. `Na_(2)SO_(4).10H_(2)O` |
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Answer» Correct Answer - D `("Molar mass of "Na_(2)SO_(4).xH_+N578(2)O)/("Molar mass of "Na_(2)SO_(4))=(100)/(44.1)` `((142+18x)u)/(142u)=(100)/(44.1)` `(142 + 18 x)u = (100)/(44.1) xx(142 u) = 321.99u` `18 x = 321.99 - 142 = 179.99` `x = (179.99)/(18)=10` `:.` Formula of crystalline salt `= Na_(2)SO_(4).10H_(2)O`. |
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| 228. |
The density of a `3 M Na_(2) S_(2) O_(3)` (sodium thiosulphate) solution is `1.25 g mL^(-1)`. Calculate: a. % by weight of `Na_(2) S_(2) O_(3)` b. Mole fraction of `Na_(2) S_(2) O_(3)` c. Molalities of `Na^(o+)` and `S_(2) O_(3)^(2-)` ions. |
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Answer» a. `Mw of Na_(2) S_(2) O_(3) = 23 xx 2 + 32 xx 2 + 16 xx 3` `= 46 + 64 + 48 = 158 g` `M = (% "by weight" xx 10 xx d)/(Mw_(2))` `3 = (% "by weight" xx 10 xx 1.25)/(150)` % by weight `= (3 xx 158)/(10 xx 1.25) = 37.92 %` b. Weight of solute `= 37.92 g` Weight of solution `= 100 g` Weight of solvent `= 100 - 37.92 = 62.08 g` `X_(2) = (W_(2)//Mw_(2))/((W_(1))/(Mw_(1))) + (W_(2))/(mw_(2)) = (37.92//158)/((62.08)/(18) + (37.92)/(158))` `= (0.24)/(3.44 + 0.24) = (0.24)/(3.68) = 0.065` c. `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (37.92 xx 1000)/(158 xx 62.08) = 3.86 m` `Na_(2)S_(2)O rarr 2 Na^(o+) + S_(2) O_(3)^(2-)` `1m` `2m` `1m` `3.86 m` `3.86 xx 2 m` `3.86 m` `m` of `Na^(o+) = 7.72 m` `m` of `S_(2) O_(3)^(2-) = 3.86 m` |
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| 229. |
Assertion (A) : The empirical mass of ethene is half of its molecular mass Reason (R) : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.A. Both A and R are true and R is the correct explanation of AB. A is true but R is falseC. A is false but R is trueD. Both A and R are false |
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Answer» Correct Answer - A Reason is the true explanation for Assertion |
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| 230. |
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M `HNO_(3)` solution ? The concentration of nitric acid is 70 % by mass. |
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Answer» No of moles `HNO_(3)=("Mass of "HNO_(3))/("Molar mass of HNO"_(3))=(W)/(("63 g mol"^(-1)))` Volume of solution in litres = 0.250 L Molarity of solution `= ("No. of moles of HNO"_(3))/("Molarity of solution in litres")` `(2.0 "mol L"^(-1))=(W)/(("63 g mol"^(-1))xx("0.250 L"))` `W=("2.0 mol L"^(-1))xx("63 g mol"^(-1))xx(0.250 L)=31.5 g` Actual mass of `HNO_(3)=(31.5g)xx(100)/(70)=45g`. |
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| 231. |
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M `HNO_(3)` solution ? The concentration of nitric acid is 70 % by mass.A. 54.0 g conc. `HNO_(3)`B. 45.0 g conc. `HNO_(3)`C. 90.0 g conc. `HNO_(3)`D. 70.0 g conc. `HNO_(3)` |
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Answer» Correct Answer - B Molarity `= ("Mass of HNO"_(3))/("Molar mass " xx " volume of solution in L")` `("2.0 mol L"^(-1))=(W)/(("63 g mol"^(-1))xx("0.250 L"))` `W=("2.0 mol L"^(-1))xx("63 g mol"^(-1))xx("0.250 L")` `=31.5g` Actual mass of `HNO_(3)=31.5gxx(70)/(100)=45.0g`. |
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| 232. |
Concentrated `HNO_(3)` is 69% by mass of nitric acid. Calculate the volume of the solution which contains `23 g` of `HNO_(3)`. (Density of concentrated `HNO_(3)` solution is `1.41 g ml^(-1))` |
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Answer» `69 g NHO_(3) = 100 g` of solution `23 g of HNO_(3) = (100)/(69) xx 33.3 g` Volume of solution `= (33.3)/(1.41) = 23.6 mL` |
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| 233. |
On heating `1.763 g` of hydrated `BaCl_(2)` to dryness, `1.505 g` of anhyrous salt remained, What is the formula of hydrate? |
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Answer» `underset(underset((208+18n))((137+71+18n)))(BaCl_(2).nH_(2)O)overset("Heat")(rarr)underset(underset((208))((137+71)))(BaCl_(2))` According to available data, `(208 + 18n)g` of `BaCl_(2).nH_(2)O` give anhydrous salt = 208 g 1.763 g of `BaCl_(2).nH_(2)O` give anhydrous salt `= ((208g))/((208+18n))xx(1.764g)` But anhydrous `BaCl_(2)` actually formed = 1.505 g or `((208))/((208+18n))xx(1.763g)=1.505g` or `(208+18n)=(208xx(1.763g))/((1.505g))=243.66` or `18n=243.66-208=35.66orn=(35.66)/(18)~~2` `:.` Formula of hydrated salt `= BaCl_(2).2H_(2)O`. |
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| 234. |
`1.375 g` of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was `1.098 g` In another experiment, `1.179 g` of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was `1.476 g`. Show that these result illustrate the law of constant composition. |
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Answer» `{:("Ist exp",,CuO=1.375gm),(,,Cu=1.098gm),(,,O=0.277gm),("IInd exp".,,Cu=1.179gm),(,,CuO=1.4476 gm),(,,O=0.2686gm):}` `(Cu)/(O) = 3.9638=4 " "(Cu)/(O)=4` In both the cases ratio of `Cu//O` is same |
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| 235. |
Statement-1 : Both solutions have equal moles of `CI^(-)` ions in given volume. Statement-2 : 2M,500ml of NaCI solution and 1 M, 500 ml of `CaCI_2` solution has equal number of `CI^(-)` ions.A. Statement-1 : is True, Statement-2 : is True, Statement-2 : is a correct explanation for Statement-1 :B. Statement-1 : is True, Statement-2 : is True, Statement-2 : is NOT a correct explanation for Statement-1 .C. Statement-1 : is Ture, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True. |
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Answer» Correct Answer - A |
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| 236. |
A solution containing `0.1` mol of a metal chloride `MCl_(x)` requires 500ml of `0.8` M `AgNO_(3)` solution for complete reaction `MCl_(x)+xAgNO_(3)rarrxAgCl+M(NO_(3))_(x)`. Then the value of x is :A. 1B. 2C. 4D. 3 |
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Answer» Correct Answer - C |
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| 237. |
Find volume of `H_(2)O` (in Litre) added to make 500ml, 1 M `NaOH(aq)` solution to `1/9` M. |
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Answer» Correct Answer - 4 |
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| 238. |
A student is asked to measure 30.0g of methanol (d = `0.7914 g//mL)` at `25^(@)C`) but has only a graduated cylinder with which to measure it. What volume of methanol should the student use to obtain the required 30.0g?A. 23.7mLB. 30.0mLC. 32.4mLD. 37.9mL |
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Answer» Correct Answer - D |
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| 239. |
An aqueous solution of `6.3 g` oxalic acid dihydrate is made up to `250 mL`. The volume of `0.1 N NaOH` required to completely neutralise `10 mL` of this solution isA. `40 mL`B. `20 mL`C. `10 mL`D. `4 mL` |
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Answer» Correct Answer - A The normality of oxalic acid dihydrate is `(6.3)/(63) xx (1)/(250) xx 100 = 0.4` `(Ew` for `(COOH))_(2) . 2H_(2)O` is 63] `N_(1) V_(1)` (acid) `= N_(2) V_(2)` (base) or `0.4 xx 10 = 0.1 xx V_(2)` or `V_(2) = 40 mL` |
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| 240. |
The molarity of a solution containing `5.0g` of NaOH in 250 mL solution is :A. `0.1`B. `0.5`C. `1.0`D. `2.0` |
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Answer» Correct Answer - B |
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| 241. |
An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is :A. 40 mLB. 20 mLC. 10 mLD. 4 mL |
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Answer» Correct Answer - A Normality of oxalic acid solution `= (6.3)/(250)xx(1000)/(63)=0.4N` Volume of NaOH solution required is calculated by applying normality equation `N_(1)V_(1) -= N_(2)V_(2)` `0.1 xx V_(1) -= 0.4 xx 10` `V_(1)=(0.4xx10)/(0.1)=40mL`. |
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| 242. |
The molarity of NaOH solution obtained by dissolving 4g of it in 250 mL solution is :A. 0.4 MB. 0.8 MC. 0.2 MD. 0.1 M |
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Answer» Correct Answer - A Molarity (M) `= ((4.0)//(40 "g mol"^(-1)))/((250//1000L))` `= 0.4 "mol L"^(-1) = 0.4 M`. |
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| 243. |
The weight of pure NaOH required to prepare 250 mL of 0.1 N solution is :A. 4 gB. 2 gC. 1 gD. 5 g |
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Answer» Correct Answer - C `("0.1 Equiv. L"^(-1))=("Mass of NaOH")/(("40 g equiv")^(-1)xx("0.2 5L"))` Mass of `NaOH = 0.1 xx 40 xx 0.25 =1g`. |
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| 244. |
Maximum number of molecules are present inA. 15 L of `H_(2)` gas at S.T.PB. 5 L of `N_(2)` gas at S.T.PC. 0.5 of `H_(2)` gasD. 10 g of `O_(2)` gas |
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Answer» Correct Answer - A (a) 15 L of `H_(2)` gas at S.T.P `= (15L)/((22.4L))xx6.022xx10^(23)=4.033xx10^(23)` (b) 5L of `N_(2)` gas at S.T.P. `= ((5L))/((22.4L))xx6.022xx10^(23)` `=1.344xx10^(23)` (c) 0.5 g of `H_(2)=((0.5g))/((2.0g))xx6.022 xx 10^(23)` `=1.505 xx 10^(23)` 10 g of `O_(2)=((10.0g))/((32.0g))xx6.022xx10^(23)` `=1.882xx10^(23)`. |
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| 245. |
The amount of chloride ion in a water sample is to be determined by adding excess silver nitrate. If 1.0g of silver chloride is precipitated, What mass of chloride ion is in the original sample? `{: ("Molar Mass", gmol^(-1)), (AgNO_(3), 169.91), (AgCl, 143.25):}`A. 0.25gB. 0.34gC. 0.50gD. 0.75g |
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Answer» Correct Answer - A |
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| 246. |
Moles of `ABC_(3)` produced in the following set tio reaction when 180gm of A, 180gm of B and 200gm of C are mixed in a container (given molar mas of A, B ,C and 20,30 and 10 respectively). `2A + 3B + 5C rightarrow A_(2)BC + B_(2)C_(3)`……………..(i) `B_(2)C_(3) + 3C rightarrow 2BC_(3)`..................(ii) `BC_(3) + A rightarrow ABC_(3)`...............(iii)A. 5B. 4C. 43741D. 20/3 |
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Answer» Correct Answer - B |
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| 247. |
Select the correct statement(s) about chemical reaction in a closed container.A. Total mass remains conserved.B. Total moles of molecules remains conserved.C. Total mass of atoms remains conservedD. Total mass of molecules may change. |
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Answer» Correct Answer - A::C::D |
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| 248. |
The concentrations of soluitons can be expressed in number of ways , viz : mass fraction of solute (or mass percent), Molar concentration (Molarity ) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each otehr i.e., knowing one concentration terms for the solution, we can find other concentration terms also. the definition of different cencentration terms are given below: Molarity : It is number of moles of solute present in one litre of the solution. Molality : It is the number of moles of solute present in one kg of the solvent. Mole fraction `=("Mole of solute")/("Moles of solute" + "Moles of solvent")` If molality of the solution is given as a, then mole fraction of the solute can be calculated by Mole Fraction `=(a)/(a+(100)/(M_("solvent"))),=(axxM_("solvent"))/((a xx M_("solvent")+1000)0` where a=molality and `M_("solvent")` =Molar mass of solvent We can change : Mole fraction `hArr` Molality `hArr` Molarity What is the mole fraction of the solute?A. 0.18B. 0.75C. 0.09D. 0.25 |
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Answer» Correct Answer - C |
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| 249. |
The concentrations of soluitons can be expressed in number of ways , viz : mass fraction of solute (or mass percent), Molar concentration (Molarity ) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each otehr i.e., knowing one concentration terms for the solution, we can find other concentration terms also. the definition of different cencentration terms are given below: Molarity : It is number of moles of solute present in one litre of the solution. Molality : It is the number of moles of solute present in one kg of the solvent. Mole fraction `=("Mole of solute")/("Moles of solute" + "Moles of solvent")` If molality of the solution is given as a, then mole fraction of the solute can be calculated by Mole Fraction `=(a)/(a+(100)/(M_("solvent"))),=(axxM_("solvent"))/((a xx M_("solvent")+1000)0` where a=molality and `M_("solvent")` =Molar mass of solvent We can change : Mole fraction `hArr` Molality `hArr` Molarity What is the molality of the above solution?A. 4.4 mB. 5.5mC. 24.4mD. None of these |
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Answer» Correct Answer - B |
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| 250. |
In a water treatment plant, `Cl_(2)` used for the treatment of water is produced from the following reaction `2KMnO_(4)+16HClrarr2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)`. If during each feed `1 L KMnO_(4)` having `79%(w//v) KMnO_(4) & 9 L HCl` with `d=1.825 g//mL & 10% (w//w) HCl `are entered `&` if that percent yield is `80%` then calculate : `(a)` amount of `Cl_(2)` produced. `(b)` amount of water that can be treated by `Cl_(2)` if `1` litre consumes `28.4 g Cl_(2)` for treatment, `(c)` calculate efficiency `eta` of the process is `eta = ("vol. of water treated")/("vol. of total feed")` |
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Answer» Correct Answer - `(a)10"mol", (b)L, (c ) 2.5` `(a)1LKMnO_(4)rarr 79%(w//v)i.e.100mL` solution contain `79 g KMnO_(4)` moles of `KMnO_(4)=(et)/(M_(w))=(79)/(158)=0.5` Molarity `(M) = (0.5)/(100)xx1000=5M` `HClrarr10%(w//w)i.e.100g` solution contain `10g HCl` `D=1.825g//mL` `V=(M)/(D)=(100)/(1.825xx1000)` Molarity `=(10xx1.825xx1000)/(36.5xx100)=5M` `2KMnO_(4)+16HClrarr2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)` `MxxV_(1)" "MxxV_(1)` `5xx1" "5xx9` `5" "5` `-" "5" "12.5` `Cl_(2)=12.5xx(80)/(100)=10`mol. (b) `2KMnO_(4)+16HClrarr2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)` `1xx(710)/(28.4)=25L` (c) `eta = ("vol of water treated")/("vol of" KMnO_(4)+HCl) = (25)/(1+9)=2.5` |
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