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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.
| 301. |
Number of electrons in 36mg of `._8^(18)O^(-2)` ions are : (Take `N_(A) = 6 xx 10^(23))`A. `1.2 xx 10^(21)`B. `9.6 xx 10^(21)`C. `1.2 xx 10^(22)`D. `1.9 xx 10^(22)` |
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Answer» Correct Answer - C |
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| 302. |
In which of the following cases, the final solution obtained will definitely be basic?A. 100 ml `0.1` M NaOH solution is mixed with 200 ml `0.1` M `H_(2)SO_(4)` solution.B. 50 ml `40%w//w` NaOH solution is mixed with 1 litre of `0.5` M `H_(2)SO_(2)` solution.C. 200 ml of `40%w//w` NaOH solution is mixed with `1.5` litre of 1 M HCl solution.D. 200 ml of `0.2` M NaOH solution is mixed with 100 ml of `0.2` M `H_(2)SO_(4)` solution. |
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Answer» Correct Answer - C |
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| 303. |
`H_(2)` gas is often used as a reducing gas. In a particular set up 17.4 gm of `MnO_(2)` on reacting with excess of hydrogen gas gives water and new oxide `Mn_(x)O_(y)` such that mass of the oxide obtained is 12.6 g. What will be value of y if x is 2. `[Mn=55]` |
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Answer» Correct Answer - 1 |
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| 304. |
Which amount of dioxygen (in grams) contains `1.8 xx 10^(22)` molecules ?A. `0.960`B. `96.0`C. `0.0960`D. `9.60` |
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Answer» Correct Answer - A `6.022 xx 10^(23)` molecules of `O_(2)` weigh = 32 g `1.8 xx 10^(22)` molecules of `O_(2)` weight `= ((32g)xx(1.8xx10^(22)))/((6.022xx10^(23)))` `=0.956~~0.96` g of `O_(2)`. |
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| 305. |
How much magnesium sulphide can be obtained from `2.00 g` of `Mg` and `2.00 g` of `S` by the reaction. `Mg + S rarr MgS`. Which is the limiting reagent? Calculate the amount of one of the reactants which remains unreacted? |
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Answer» `Mg + S rarr MgS` `24 g 32 g 32 + 24 = 56 g` `32 g` of `S` in obtianed from `24 g` of `Mg` `2 g` of `S` in obtained from `= (24)/(32) xx 2 = 1.5 g of Mg` So, `S` is completely consumed and `Mg` is left. Hence, `S` is the limiting reagent. Therefore, the amount of product, i.e., `MgS`, will be determined from `S` and not from `Mg`. `32 g` of `S = 56 g of MgS` `2 g of S = (56)/(32) xx 2 = 3.5 g of MgS` Amount of `Mg` unreacted `= 2 - 1.5 = 0.5 g` |
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| 306. |
In the reaction: `A+B_(2) rarr AB_(2)` Find out in which option(s) `B_(2)` is limiting reagent. [Molar mass : `M_(A)=10 g "mole"^(-1), M_(B)=20 g "mol"^(-1)`]A. 300 atoms of A + 200 molecules of `B_(2)`B. 2 mole A+3 gram-atom of BC. 100 mole A atoms +100 mole B atomsD. 5 gram-atom of A + 12.5 gram molecule of `B_(2)` |
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Answer» Correct Answer - A::B::C |
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| 307. |
A mixture of `O_(2)` and gas `"Y"` `( mol. wt. 80)` in the mole ratio `a:b` has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio `b:a` under identical conditions ? ( gases are )A. 40B. 48C. 62D. 72 |
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Answer» Correct Answer - 4 Let mole fraction of `O_(2)` is `x` `40=32xx x+80(1-x)" "`or `" " x=5//6` `a+b=x:(1-x)=(5)/(6):(1)/(6)` When ratio is changed `M_("mixture")=32xx(1)/(6)+80xx(1)/(6)=72` |
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| 308. |
What is the minimum amount of `Fe(NH_(4))_(2)(SO_(4))_(2) H_(2)O (392)` needed for the synthesis of 10.0 g of `K_(3)Fe(C_(2)O_(4))_(3)(437)`? Give your answer to the nearest integer. |
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Answer» Correct Answer - 9 |
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| 309. |
The normality of 0.3 M phosphorus acid `(H_(3)PO_(3))` solution is :A. 0.1B. 0.9C. 0.3D. 0.6 |
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Answer» Correct Answer - D `H_(3)PO_(3)` is a dibasic acid `({:(" O"),(" ||"),(HO-P-H),(" |"),(" OH"):})` its basicity = 2 Normality = Molarity `xx` Basicity `= 0.3 xx 2 = 0.6 N`. |
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| 310. |
From 200 mg of `CO_(2),10^(21)` molecules are removed. How many moles of `CO_(2)` are left ? |
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Answer» Step I : No. of molecules in 200 mg of `CO_(2)` Molar mass of `CO_(2)=12+32 = 44` g 44 g of `CO_(2)` contain molecules `= 6.022 xx 10^(23)` 200 mg or 0.2 g of `CO_(2)` contain molecules `= (6.022xx10^(23))/((44g))xx(0.2g) = 2.74 xx 10^(21)` Step II. No. of molecules of `CO_(2)` left Total no. of `CO_(2)` molecules `= 2.74 x 10^(21)` No. of `CO_(2)` molecules removed `= 10^(21)` No. of `CO_(2)` molecules left `= 2.74 xx 10^(21) - 10^(21) = 1.74 xx10^(21)` Step III. No. of moles of `CO_(2)` left `6.022 xx 10^(23)` molecules of `CO_(2)=1` mol `1.74xx10^(21)` molecules of `CO_(2)=(1 mol)/(6.022xx10^(23))xx1.74xx10^(21)=2.89xx10^(-3)` mol. |
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| 311. |
If `10^(21)` molecules are removed from 200 mg of `CO_(2)`, the number of moles of `CO_(2)` left will be ?A. `2.88 xx 10^(-3)`B. `1.66 xx 10^(-3)`C. `4.54 xx 10^(-3)`D. `1.66 xx 10^(-2)` |
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Answer» Correct Answer - A No. of molecules in 200 mg of `CO_(2)` `= (6.022 xx 10^(23))/(44)xx0.2=2.73xx10^(21)` No. of molecules removed `= 10^(21)` No. of molecule left `= (27.73 xx 10^(21) - 1 xx 10^(21))` `= 1.73 xx 10^(21)` molecules No. of moles left`= (1.73xx10^(21))/(6.022xx10^(23))` `=2.87xx10^(-3)` |
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| 312. |
How many moles of water are produced by the complete comustion of `14.4` "g of" `C_5H_12`? `C_5H_12+8O_2to5CO_2+6H_2O`A. `0.200`B. `0.600`C. `1.20`D. `2.40` |
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Answer» Correct Answer - C |
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| 313. |
Consider following vessel of given dimensions Cubical vessel is filled completely with `2M H_(2)SO_(4)` solution , whereas spherical and cylindrical vessels are empty (Tale `pi`=3 ) (Atomic mass of Br =80) The acid left in cylindrical container and its molarity will be :A. `H_(2)SO_(4), 2M`B. `H_(2)SO_(4),(2M)/(3)`C. `HCl ,(2M)/(3)`D. `HCl,(4)/(3) M` |
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Answer» Correct Answer - D |
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| 314. |
Consider following vessel of given dimensions Cubical vessel is filled completely with `2M H_(2)SO_(4)` solution , whereas spherical and cylindrical vessels are empty (Tale `pi`=3 ) (Atomic mass of Br =80) Total centent of cubical vessel is placed in cylindrical vessel and it is further filled completely with 1 M `BaCl_(2)` solution. Molarity of `SO_(4)^(2-)` ions in aqueous solution will be :A. 1 MB. 0 MC. 0.66 MD. 0.33 M |
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Answer» Correct Answer - B |
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| 315. |
Consider following vessel of given dimensions Cubical vessel is filled completely with `2M H_(2)SO_(4)` solution , whereas spherical and cylindrical vessels are empty (Tale `pi`=3 ) (Atomic mass of Br =80) Now total content of cuylindrical container is placed in a spherical container and if it is further filled completely with 32.4 % w/v HBr solution. Molarity of `H^(+)` in final solution will be:A. `(7)/(4)` MB. 2MC. `(3)/(2)` MD. `(5)/(2)`M |
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Answer» Correct Answer - B |
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| 316. |
`20 mL of 0.2MAl_(2)(SO_(4))_(3)` mixed with `20 mL` of `0.6M BaCl_(2)`. Calculate the concentration of each ion in solution. |
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Answer» `Al_(2) (SO_(4))_(3) + BaCl_(3) + BaCl_(2) rarr BaSO_(4) darr + AlCl_(3)` mEq before `20 xx 0.2 xx 6` `20 xx 0.6 xx 2` 0 0 mEq after 0 0 24 24 mixing mEq `= M xx V` (in `mL`) `M xx` Valency `:. [Al^(3+)] = (24)/(40 xx 3) = 0.2 M` Total volume `= 20 + 20 = 40 mL` `[Cl^(ɵ) = 0.6 M` No concentration of `Ba^(2+)` or `SI_(4)^(2-)` in the solution since `BaSO_(4)` gets precipitated. |
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| 317. |
How much `BaCl_(2)` would be needed to make `250 mL` of a solution having same eoncentration of `Cl^(o+)` as the one containing `3.78 g` of `NaCl` per `100 mL`. |
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Answer» `N_(NaCl) = (3.78 xx 1000)/(58.5 xx 100)` `= 0.646` Let `W g` of `BaCl_(2)` is dissolved in `250 mL`, then `N_(BaCl_(2)) = (W)/((208)/(2) xx (250)/(1000)) = 0.038 W` `:. [Cl^(ɵ)]` in both is same `:. N_(NaCl) = N_(BaCl_(2))` `:. 0.646 = 0.085 W` `: W = 16.08 g` |
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| 318. |
`A` chloride mixture is prepared by grinding together pure `BaCl_(2).2H_(2)O, KCl` and `NaCl`. What is the smallest and largest volume of `0.15 M AgNO_(3)` solution that may be used for complete precipitation of chloride from a `0.3 g` sample of the mixture which may contain any one or all of the constituents ? |
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Answer» Correct Answer - `16.38mL,34.18L` Smallest volume of `AgNO_(3)` would be required when the entire mass is due to highest molecular weight constituent. Hence, for smallest volume, the whole mass should be of `BaCl_(2). 2H_(2)O` `m` mol of `BaCl_(2)2H_(2)O=(0.3)/(244)xx1000=1.229m` mol `m` mol of `AgNO_(3)`required `=2xx1.229 = 2.458` Volume of `AgNO_(3)` required `=(2.458)/(0.15)=16.38mL` Largest volume of `AgNO_(3)` would be required when entire mass is due to lowest molecular weight constitute, i.e., `NaCl`. `m` mol of `NaCl = (0.3)/(58.5) xx1000 = 5.128 = m` mol of `AgNO_(3)` required implies Volume of `AgNO_(3)` required `=(5.128)/(0.16) = 31.18ml. ("largest")` |
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| 319. |
`A` mother cell disintegrate into sixty identical cells and each daughter cell further disintegrate into `24` smaller cells. The smallest cells are uniform cylindrical in shape with diameter of `120 Å` and each cell is `6000 Å` long. Determine molar mass of the mother cell if density of the smallest cell is `1.12 g//cm^(3)` : |
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Answer» Correct Answer - `6.6 xx 10^(10)g` Volume of smallest cell `pir^(2)|=pi(60xx10^(-8)cm)^(2)` `(6000xx10^(-8)cm) = 6.785xx10^(-17)cm^(3)` mass of one smallest cell `=7.6xx10^(17) g` implies Molar mass of mother cell `=7.6xx10^(-17)xx24xx60xx6.023xx10^(23)=6.6xx10^(10)"amu"` |
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| 320. |
A hydrate of magnesium iodide has a formula `Mgl_(2).xH_(2)O.A 1.055 g` sample is heated to a constant weight of `0.695 g`. What is the value of `x` ?A. `2`B. `4`C. `6`D. `8` |
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Answer» Correct Answer - D `(W_("sample"))/(W_("residuce")) = (M_(Mgl_(2).xH_(2)O))/(M_("residuce"(Mgl_(2))))` `(1.055)/(0.695) = (278+18x)/(278)` x = 8` |
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| 321. |
Number of neutrons in 5.5gm `T_(2)O` (T is `._1H^(3))` are.A. `0.25 N_(A)`B. `2.5N_(A)`C. `3N_(A)`D. None of these |
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Answer» Correct Answer - C No. of neutrons `=(5.5)/(22)xxN_(A)xx12 = 3 N_(A)` |
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| 322. |
Which of the following terms are unitless ?A. MolalityB. MolarityC. Mole fractionD. Mass percent |
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Answer» Correct Answer - C::D Both mole fraction and mass percent are unitless |
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| 323. |
A faculty of KOTA Institute who has five classes per day, after taking the fourth class and before going for next class, drinks 500 ml of `90%(w//v)` glucose solution. The number of glucose molecules taken by him is `=Qxx10^(23)`. Find the value of Q. `(N_(A)=6.0xx10^(23))`A. 9B. `6.5`C. `4.5`D. `1.5` |
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Answer» Correct Answer - D |
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| 324. |
`H_(2)O_(2)` solution of 45.4 V at STP . In a hospital of Kota, patient under artifical respiration take s200ml `O_(2)` per min at 1 atm and 273 K and a cylinder last for 2.8 hours . After that it cannot be used for respiration though it still contains `H_(2)O_(2)` . [Assume volume of solution and rate of decomposition remain constant] Initial moles of `H_(2)O_(2)` in a cylinder are:A. 12B. 16C. 8D. 24 |
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Answer» Correct Answer - A |
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| 325. |
Density of a gas relative t air is `1.17`. Find the mol. Mass of the gas `[M_(air)=29g//mol]` |
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Answer» Correct Answer - `33.9` `(M_("gas"))/(M_("air"))=1.17impliesM_(gas)=1.17xx29=33.93gm` |
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| 326. |
One type of artificial diamond (commonly called YAG for yttrium aluminium garnet) can be represented by the formula `Y_(3)Al_(5)O_(12)` `(a)` Calculate the weight percentage composition of this compound. `(b)` What is the weight of yttrium present in a `200 -` carat `YAG` if `1` carat `-200mg` ? `(Y=89),Al=27)` |
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Answer» Correct Answer - `(a) Y=44.95%,Al=22.73%,O=32.32%" "(b)17.98g` `Y_(3)A_(5)O_(12)` `200xx200xx10^(-3)` `(a)y=44.95%, " "Al=22.73%," "O=32.32%` `(b) 17.98 gm` |
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| 327. |
What is the percentage of nitrogen in an organic compound `0.14 g` of which gave by Dumas method `82.1 c.c`. Of nitrogen collected over water at `27^(@)C` and at a barometric pressure of `774.5 mm` ? `("aqueous tension of water at" 27^(@)C "is" 14.5 mm)` |
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Answer» Correct Answer - `66.7%` `n_(N_(2))=(((774.5-14.5))/(760)xx(82.1)/(1000))/(0.081xx300)=3.3786xx10^(-3)` mole `m_(N_(2))=0.0946gm` `% N_(2)=(0.0946)/(0.14)xx100=66.7%` |
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| 328. |
The number of atoms in 0.1 mole of triatomic gas is :A. `6.026 xx 10^(22)`B. `1.806 xx 10^(23)`C. `3.6 xx 10^(23)`D. `1.8 xx 10^(22)` |
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Answer» Correct Answer - B No. of atoms of `= N_(A) xx` No. of moles `xx 3` `= 6.022 xx 10^(23) xx 0.1 xx 3` `=1.806 xx 10^(23)`. |
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| 329. |
Calculate the molar mass of a compound in the Dumas method at `100^(@)C` for which volume of experimental container was 452 ml and the pressure was 745.1 torr. The difference in mass between the empty container and the final measurement was 1.129 gm. |
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Answer» `n=(PV)/(RT)=(745.1)/(760)xx(452xx10)/(0.0821xx373)=0.01448` mol molar mass (M)`=(1.129)/(0.01448)=78.0` gm/mol |
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| 330. |
The density of a gaseous element is 5 times that of oxygen under similar conditions. If the molecule is triatomic, what will be its atomic mass ? |
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Answer» Molecular mass of oxygen `= 32 u` Density of oxygen `= (32)/(2) =16` u Density of gaseous element `= 16 xx 5 = 80` u Molecular mass of gaseous element `= 80 xx 2 = 160` u Atomicity of the element = 3 `:.` Atomic mass of the element `= ("Molar mass")/("Atomicity")=(160)/(3)=53.33` u. |
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| 331. |
`27.6gK_(2)CO_(3)` was treated by a series of reagents so as to convent all of its carbon to `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)` Calculate the weight of the product [mol.wt. of `K_(2)CO_(3)=138` and mol. Wt. of `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)=698]` . |
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Answer» Here we have no knowledge about series of chemical reactions but we know initial reactant and final product, accordingly. `K_(2)CO_(3) underset("Steps")overset("Several")to K_(2)Zn_(3)[Fe(CN)_(6)]_(2)` Since C atoms are conserved, applying POAC for C atoms, moles of C in `K_(2)CO_(3)` =moles of C in `K_(2)Zn_(3) [Fe(CN)_(6)]_(2)` `1 xx ` moles of `k_(2)CO_(3)=12 xx` moles of `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)` `(because ` 1 mole of `K_(2)CO_(3)` contains 1 moles of `C)` `("wt. of" K_(2)CO_(3))/("mol. wt. of" K_(2)CO_(3))=12xx("wt. of the product")/("mol. wt. of product")` wt. of `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)=(27.6)/(138)xx(698)/(12)=11.6 g` |
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| 332. |
A `120gm CaCO_(3)` sample having inert impurities on heating produced `56gm` of residue. Find `%` punity of sample . |
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Answer» `{:("Balanced Rxn". ,CaCO_(3)(s),overset(Delta)rarr,CaO(s),+,CO_(2)(g)),(,,1mol,,56gm,),(,,=100ml,=mol,,):}` `%` purity `=(100)/(120) xx 100 = 83.3 %` . |
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| 333. |
For the reaction `A(s)+2B(g)rarr3C(g)+D(g)` Molar mass `[g mol^(-1)] A = 100, B = 50, C = 60, D = 20` (a) How many grams of `C` are produced by reaction of `250g A` ? [Hint : Mass relation] (b) What mass of `B` reacts to give `500g D` (c) How many grams of A will produce `11.2L` of `C` at `STP` ? . |
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Answer» `{:(A(s),+,2B(g),rarr,3C(g),+,D(g)),((250)/(100) = 2.5 mol,,"excess",,7.5mol,,):}` `= 7.5 xx 60 g` `= 450gm` `{:((b)A(s),+,2B(g),rarr,3C(g),+,D(g)),(,,50 mol,,(500)/(20) = 25 mol,,):}` `= 50 xx 50` `= 2500 gm` (c ) [Hint : mass-volume relation] `{:(A(s),+,2B(g),rarr,3C(g),+,D(g)),(,,(0.5)/(3) mol,,(11.2)/(22.4) = 0.5 mol,,):}` `= (0.5)/(3) xx 100` `= 16.67 gm` |
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| 334. |
Calculate no. of carbon and oxygen atoms present in 11.2 litres of `CO_(2)` at N.T.P |
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Answer» Step I : No. of `CO_(2)` molecules in 11.2 litres `22.4` litres of `CO_(2)` at N.T.P = 1 gram mol 11.2 litres of `CO_(2)` at N.T.P `= ((1 "gram mol"))/((22.4 "litres"))xx(11.2"litres")=0.5` gram mol Now 1 gram mole of `CO_(2)` contain molecules `= 6.022 xx 10^(23)` `:. 0.5` gram mole of `CO_(2)` contain molecules `= 6.022xx10^(23)xx0.5=3.011xx10^(23)` Step II. No. of carbon and oxygen atoms in `3.011 xx 10^(23)` molecules of `CO_(2)` 1 molecule of `CO_(2)` contains carbon atoms = 1 `:. 3.011 xx 10^(23)` Similarly, 1 molecule of `CO_(2)` contains oxygen atoms = 2 `:. 3.011 xx 10^(23)` molecules of `CO_(2)` will contain oxygen atoms `= 2 xx 3.011 xx 10^(23) = 6.022 xx 10^(23)` atoms. |
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| 335. |
What is the mass of carbon present in 0.5 moles of `K_(4)[Fe(CN)_(6)]` ? |
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Answer» Correct Answer - 36 g 1 mole of `K_(4)[Fe(CN)_(6)]` contain carbon = 6 g atoms 0.5 mole of `K_(4)[Fe(CN)_(6)]` contain carbon = 3g atoms Mass of carbon in grams `= (12g)xx(("3 g atoms"))/(("1g atom"))=36g` . |
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| 336. |
Calculate number of gm ions present in an aqueous solution containing 369 gm of `K_(2)SO_(4).(NH_(4))_(2)SO_(4). 24H_(2)O` if the salt undergoes complete dissocation into ions and water does not dissociate. |
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Answer» Correct Answer - 3 |
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| 337. |
Calculate total number of `SO_(2)` molecules in a sample having 32 milligrams of the `SO_(2)` gas, `1.4xx10^(20)` number of `SO_(2)` molecules, 0.8 ml of `SO_(2)` gas at 6 atm and 300 K. [Given: `N_(A)=6xx10^(23)`, `R=0.08 atm L//mol K]` [Express your answer in terms of multiple of `10^(20)` and then round off to nearest integer for e.g. if your answer is `6.2xx10^(20)` fill 6 in OMR after rounding off.] |
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Answer» Correct Answer - 6 |
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| 338. |
Statement-1 : When 1 mole of `NH_3` and 1 mole of `O__(2)` are made to react, all the `NH_(3)` may be consumed, if reactions is : `NH_(3)(g)+O_(2)(g)toNO(g)+H_(2)O(l)` Statement-2 : Oxygen is limiting reagent.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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| 339. |
Certain metals have a tendency to form compounds which have a "comple structure" and are known as complex. M is a metal with such tendencies and is forming compound like `[M(NH_(3))_(5)Br[Br_(2)`. If it is known that solution of these furnish only those ions which are outside the co-ordination sphere "[ ]" (the bracketed part), then calculate the weight (in g) of AgBr ppt obtained when 1000 g solution of the complex compound containing 40 % by wt. of the complex compound is reacted with 1000 g solution of `AgNO_(3)` containing 17 % `AgNO_(3)` by weight. [Atomic weight of M = 75] |
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Answer» Correct Answer - 188 |
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| 340. |
`A+BtoA_3B_2`(unbalanced) `A_3B_2+C to A_3B_2C_2` (unbalanced) Above two reactions are carried out by taking 3 moles each of A and B and one mole of C.Then which option is/are correct ?A. 1 mole of `A_3B_2C_2` is formedB. `1/2` mole of `A_3B_2C_2` is formedC. 1 mole of `A_3B_2` is formed from first reactionD. `1/2` mole of `A_3B_2` is left finally. |
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Answer» Correct Answer - B::C::D |
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| 341. |
Which is/are correct statement about 1.7 gm of `NH_3` ?A. It contains 0.3 mol H-atomB. It contains `2.408xx10^(23)` atomsC. Mass % hydrogen is 17.65%D. Vapour density of `NH_3` is 17 |
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Answer» Correct Answer - A::B::C |
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| 342. |
Water is the working fluid used in Kota thermal power plant for generating electricity. Coal is combusted for generating heat as per reaction, `C + O_(2) rarr CO_(2). 0.01 %` of the released `CO_(2)` gas is absorbed in water and gets coverted to weak acid, `H_(2)SO_(3)` which dissociated to give `H^(+)` as `H_(2)CO_(3) rarr 2H^(+) + CO_(3)^(2-)`. The percentage dissociattion of acid is 5 %. Assume no ionisation of water. From this information answer the questions. if in a certain application `[H^(+)]` concentration can maximum be `10^(-5)M`, then, (P) Calculate maximum moles of `H^(+)(x)` and `CO_(3)^(2-)(y)` in the water water if `10^(9)` litres of `H_(2)O` is used (Q) Calculate maximum moles of carbon `(z)` which can be burnt so that water remains fit to be used Hence, write the value of `(xy^(2)//z)` |
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Answer» Correct Answer - 250 |
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| 343. |
Where do we use the words mole and mol ? |
| Answer» In the text part, we use the word mole while as a unit, we call it mol | |
| 344. |
Which of the following will occupy greater volume under the similar conditions of pressure and temperature?A. 6gm oxygenB. 0.98 gm hydrogenC. 5.25 gm of nitrogenD. 1.32 gm of helium |
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Answer» Correct Answer - B |
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| 345. |
Does one gram mole of a gas occupy 22.4 L under all conditions of temperature and pressure ? |
| Answer» No, one gram mole of a gas occupies 22.4 L only under N.T.P or S.T.P conditions i.e., at 273 K temperature and under 760mm pressure. If these conditions are not used, then the volume is not 22.4 L. | |
| 346. |
A cylinder of compressed gas contains nitrogen and oxygen in the molar ratio of `3:1`. If the cylinder contains `2.5 xx 10^(4)g` of oxygen, what is the total mass of the gas in the mixture ? |
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Answer» Number of moles of oxygen `= ("Mass of oxygen")/("Molar mass of oxygen")=((2.5xx10^(4)g))/((32.0" g mol"^(-1)))=781.25` mol Number of mass of nitrogen `= 3xx781.25=2343.75` mol Mass of nitrogen `= (2343.75 " mol")xx(28 " g mol"^(-1))` `=65625g=6.5625xx10^(4)`g Total mass of gas (oxygen + nitrogen) `= (2.5 xx 10^(4)+6.5625xx10^(4)g)=9.0625xx10^(4)g`. |
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| 347. |
In a bank, there are as many coins as the number of molecules in `1.6 mug` of `CH_(4)`. How many coins are there in the bank ? |
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Answer» Numbe of moles present `= ("Mass of methane")/("Molar mass of methane")` `=((1.6xx10^(-6)g))/((16.0"g mol"^(-1)))=10^(-7) mol" "(1.0 mu g = 10^(-6) g)` Number of molecules present `= (10^(-7))xx(6.022xx10^(23))=6.022xx10^(16)` `:.` Number of coins in the bank `= 6.022 xx 10^(16)`. |
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| 348. |
A sample of `MgSO_(4)` has `6.023 xx 10^(20)O` atoms What is mass of `Mg` in sample ? . |
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Answer» In `MgSO_(4) ("mass of O")/("mass of Mg") = (16 xx 4)/(24) = (8)/(3)` = constant Mass of `6.023 xx 10^(20)` O-atoms `= (6.023 xx 10^(20))/(6.023 xx 10^(23)) xx 16 = 16 xx 10^(-3) gms` `rArr (8)/(3) = (16 xx 10^(-3))/("mass of Mg") rArr` Mass of Mg `= 6 xx 10^(-3)gms` . |
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| 349. |
If the percentage of water of crystallization in `MgSO_(4) .x H_(2)O` is `13%`. What is the value of `x` :A. `1`B. `4`C. `5`D. `7` |
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Answer» Correct Answer - A `%` by wt of `H_(2)O` `("wt.of" H_(2)O)/("Total wt.of compound")xx100` `13 = (18x)/(18x + 120) xx 100` `x = 1` . |
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| 350. |
Calculate the weight of `90%` pure sulphuric acid to neutralize `5g` caustic soda.A. `19.19`B. `6.81`C. `11.11`D. None of these |
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Answer» Correct Answer - B Writing the equation for the above reaction `{:(2NaOH+,,H_(2)SO_(4),rarr,Na_(2)SO_(4)+2H_(2)O),(2 mol es,,1 mol e,,),(or 80g,,,or 98g,):}` No. of moles in `5 g` caustic soda `=(5)/(40)` `2` mole of `NaOH` need for neutralization `=1 `mole of `H_(2)SO_(4)` `:. (5)/(40)` mole of `NaOH` need for neutralization `=(1)/(2)xx(5)/(40)=(1)/(16)mol.` Amount of `H_(2)SO_(4)` in `g=(1)/(16)xx98g H_(2)SO_(4)=6.13g` Amount of `90%` pure `H_(2)SO_(4)=6.13xx(100)/(90)=6.81` |
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