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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.
| 151. |
How many `g` of element are present in `35,125 g` atom of `Si`. `("Given at". `wt` "of" Si = 28.)` |
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Answer» Correct Answer - `983.5g` of `Si` `35.125xx28=983.5 gm` |
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| 152. |
The density of liquid (mol.wt. `= 70`) is `1.2 g mL^(-1)`. If `2 mL` of liquid contains `35` drops, the number of molecules of liquid in one drop are:A. `((1.2)/(35))N_(A)`B. `(1)/(1.2)((1)/(35))^(2)N_(A)`C. `(1.2)/((35)^(2))N_(A)`D. `1.2N_(A)` |
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Answer» Correct Answer - 3 Mass of one drop `=(2)/(35)xx1.2=(2.4)/(35)g` Moles of liquid in one drop `=(2.4)/(35xx70)=(1.2)/(35^(2))` `:. ` Molecules `=(1.2)/(35^(2))xxN_(A)` |
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| 153. |
Sulphur exist in different allotropic forms like `S_(2),S_(6)` and `S_(8)` etc. If equal moles of these three forms are taken in separate containers, then the ratio of number of atoms present in them respectively is `:`A. `1:3:4`B. `1:1:1`C. `12:4:3`D. `4:3:1` |
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Answer» Correct Answer - A `{:(,S_(2),S_(6),S_(8)),("mole",n,n,n),("molecule",nxxN_(A),nxxN_(A),nxxN_(A)),("atoms",nxxN_(A)xx2,nxxN_(A)xx6,nxxN_(A)xx8):}` `:. ` Ratio of atoms of `S=2:6:8=1:3:4` |
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| 154. |
Find the normality of `H_(2) SO_(4)` having 50 milli equivalents in 2 litres. |
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Answer» Correct Answer - B `N = (g "equivalent")/(V_(sol)("in" L)) = (mEq)/(V_(sol)("in" mL)) = (50)/(2000) = 0.025 N` |
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| 155. |
1.0 g of a mixture of carbonates of calcium and magnesium gave `240 cm^(3)` of `CO_(2)` at N.T.P Calculate the percentage composition of the mixture. |
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Answer» Correct Answer - `CaCO_(3)=62.5 %, MgCO_(3)=37.5 %` Let the mass of `CaCO_(3)` in the mixture = x g `:.` The mass of `MgCO_(3)` in the mixture `= (1-x)g` Step I. Volume of `CO_(2)` evolved from `CaCO_(3)` `underset(40+12+48+100g)(CaCO_(3)(s))overset("Heat")(rarr)CaO(s)+underset(22400"cc")(CO_(2)(g))` 100 g of `CaCO_(3)` evolve `CO_(2)` upon heating = 22400 cc xg of `CaCO_(3)` evolve `CO_(2)` upon upon heating `= (22400)/(100) xx x c c = (224x)c c`. Step II. Volume of `CO_(2)` evolved from `MgCO_(3)` `underset(24+12=48=84g)(MgCO_(3)(s))overset("Heat")(rarr)MgO(s)+underset("22400cc")(CO_(2)(g))` 84 g of `MgCO_(3)` evolve `CO_(2)` upon heating = 22400 cc `(1-x)`g of `MgCO_(3)` evolve `CO_(2)` upon heating `= 22400 xx ((1-x))/(84) c c. = 266.6 xx (1-x)c c` Step III. Percentage composition of the mixture According to available data : `224 x + 266.6 (1-x)=240` `224 x + 266.6 - 266.6 x = 240` `- 42.6 x = -26.6 or x = (26.6)/(42.6)=0.625` `:.` Percentage of `CaCO_(3)` in the mixture `= 0.625 xx 100 = 62.5 %` Percentage of `MgCO_(3)` in the mixture `= 100 - 62.5 = 37.5 %`. |
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| 156. |
Whay is it necessary to balance a chemical equation ? |
| Answer» A chemical equation has to be balanced in order to satisfy the law of conservation of mass. According to the law, there is no change in mass when the reactants change into the products. Therefore, the chemical equation has to be balanced. | |
| 157. |
Hydrogen catches fire easily and oxygen helps in combusion. Why is water used for extinguishing fire ? |
| Answer» Since water `(H_(2)O)` is a compound, its properties are quite different from those of hydrogen and oxygen. | |
| 158. |
Are the atomic masses of some elements actually fractional ? |
| Answer» No, the actual masses are not fractional. It is the average of the atomic masses of the isotopes of an element which maybe fractional. For example, the two isotopes of chlorine have atomic masses 35u and 37u respectively. The average atomic mass of chlorine is 35.5 u. | |
| 159. |
Whay is air not always regarded as homogeneous mixture ? |
| Answer» Air which contains certain suspended particles such as dust particles is a hetergeneous mixture and not homogeneous mixture. | |
| 160. |
Which of the following has maximum mass ? (i) 0.1 mole of `KNO_(3)` (ii) `1 xx 10^(23)` molecules of `CH_(4)` (iii) `112 cm^(3)` of hydrogen at S.T.P ? |
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Answer» Correct Answer - 0.1 mole of of `KNO_(3)` has maximum mass of 10.1 g (i) Formula mass of `KNO_(3)=1 xx 39 + 1 xx 14 + 3 xx 16 = 101` 1 mole of `KNO_(3)=101 g` 0.1 mole of `KNO_(3)=(("0.1 mol"))/(("1.0 mol"))xx(101g)=10.1g` (ii) `6.022 xx 10^(23)` molecules of `CH_(4)=16 g` `1 xx 10^(23)` molecules of `CH_(4)=((1xx10^(23)"molecules"))/((6.022xx10^(23)"molecules"))xx(16g)=2.657g` (iii) `22400 cm^(3)` of `H_(2)` at S.T.P = 2g `112 cm^(3)` of `H_(2)` at S.T.P `= (("112 cm"^(3)))/(("22400 cm"^(3)))xx2g=0.01g` Thus, 0.1 mole of `KNO_(3)` has maximum mass. |
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| 161. |
How many moles of methane are required to produce 22g of `CO_(2)` on combustion ? |
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Answer» The balanced equation for the combustion reaction is : `underset("1 mol")(CH_(4)(g))+2O_(3)(g)rarrunderset("1 mol")(CO_(2)(g))+2H_(2)O(l)` No. of moles in 22 g of `CO_(2)=("Mass of "CO_(2))/("Molar mass")=((22g))/((44"g mol"^(-1)))=0.5` mol 1 mole of `CO_(2)` is produced from methane = 1 mol 0.5 mole of `CO_(2)` is produced from methane = 0.5 mol. |
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| 162. |
Calculate the weight of carbon monoxide having the same number of oxygen atoms as are present in 22g of carbon dioxide |
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Answer» Step I : No. of the oxygen atoms in 22 g of `CO_(2)` Molar mass of `CO_(2)+12 + 2 xx 16 = 44 g` 44g of `CO_(2)` represent = 1 mol 22g of `CO_(2)` represent `= (1mol) xx((22g))/((44g))=0.5 mol` Now, 1 mole of `CO_(2)` contain oxygen atoms `= 2xx 6.022 xx 10^(23)` `0.5` moles of `CO_(2)` contain oxygen atoms `= 2xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(23)` atoms Step II. Weight of carbon monoxide Molar mass of `CO=12+16 = 28` g By definitioon, `6.022 xx 10^(23)` atoms (or 1 mole oxygen atoms) are present in CO = 28 g. |
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| 163. |
A chemist wishes to prepare `6.022 xx 10^(24)` molecules of sulphur dioxide. How many gram-atoms of sulphur does the need ? |
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Answer» Correct Answer - 10 gram atoms `6.022 xx 10^(23)` molecules of `SO_(2)=1` gram atom of `SO_(2)=1` gram atom of S `6.022 xx 10^(24)` molecules of `SO_(2)=10` gram atom of `SO_(2)=10` gram atoms of S. |
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| 164. |
Carbon reacts with oxgen forming carbon monoxide and/or carbon dioxide depending an availablity of oxygen. Find moles of each product obtained when `160gm` oxygen reacts with (a) 12 g carbon (b) 120 g carbon (C ) 72 g carbon. |
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Answer» (a) `C +(1)/(2)O_(2) to CO" "` [initially use a reaction lesser amount of oxygen] `{:(t=0,1 "mol",5 "mol",,),(t=oo,0,5-0.5,=,1 "mol"),(,(LR),4.5"mol",,):}` Since CO & `O_(2)` are left `CO_(2)` is formed `CO + (1)/(2) O_(2) to CO_(2)` `{:(t=0,1 "mol",4.5 "mol",,0),(t=oo,0,4 "mol",=,1 "mol"):}` At end 1 mole `CO_(2)` & no CO present (b) `C + (1)/(2)O_(2) to CO` `{:(t=0,10 "mol",5 "mol",,0),(t=oo,0,0 ,,10 "mol"):}` (c) `C+(1)/(2)O_(2) to CO` `{:(t=0,10 "mol",5 "mol",,0),(t=oo,0,2"mol" ,,6 "mol"),(,[LR],,,):}` `CO+(1)/(2)O_(2) to CO_(2)` `{:(t=0,6 "mol",2 "mol",,0),(t=oo,2 "mol",0[LR] ,,4 "mol"):}` At end [2 mol CO + 4 mol `CO_(2)`]left |
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| 165. |
Cisplation is used an anticancer agent for the treatment of solid tumors, and its prepared as follows: `underset({:("Potassium tetra"),("Chloro platinate"):})(K_(2)[PtCl_(4)]) + 2NH_(3) rarr underset("Ciplatin")([Pt(NH_(3))_(2) Cl_(2)]) + 2KCl` Given `83.0 g` of `K_(2) [PtCl_(4)]` is used with `83.0 g` of `NH_(3)`. Atomic weights: `K = 39 , Pt = 415, Cl = 35.5 N = 14]` The number of mol of excess reactant isA. 4.68B. 4.78C. 4.58D. 4.48 |
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Answer» Correct Answer - D Excess of `NH_(3)` unreacted `= 4.88 = 4.48 "mol"` |
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| 166. |
A 1.0 gram sample of which substance contains the largest number of molecules?A. `COCl_(2)`B. `CS_(2)`C. `CH_(3)Cl`D. `C_(2)H_(2)F_(2)` |
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Answer» Correct Answer - C |
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| 167. |
Calculate the volume at N.T.P occupied by (i) 14 g of nitrogen (ii) 1.5 gram moles of carbon dioxide (iii) `10^(21)` molecules of oxygen. |
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Answer» (i) Volume of 14 g of nitrogen at N.T.P Molar mass of nitrogen `(N_(2))=28` g 28 g of `N_(2)` at N.T.P. occupy volume `= 22.4 L` 14 g of `N_(2)` at N.T.P occupy volume `= ((22.4L))/((28g))xx(14g)=11.2L` (ii) Volume of 1.5 gram moles of carbon dioxide at N.T.P 1.0 gram mole of `CO_(2)` at N.T.P occupy volume `=22.4L` 1.5 gram moles of `CO_(2)` at N.T.P occupy volume `=22.4 xx 1.5 = 33.6`L (iii) Volume of `10^(21)` molecules of oxygen at S.T.P We know that, 22.4 L is the volume of `6.022 xx 10^(23)` molecules of a gas at N.T.P Thus, `6.022 xx 10^(23)` molecules of `O_(2)` at N.T.P occupy volume = 22.4 L `10^(21)` molecules of `O_(2)` at N.T.P occupy volume `= ((22.4L)xx10^(21))/(6.022xx10^(23))=3.72xx10^(-2)L` `=3.72xx10^(-2)xx10^(3)mL=37.2mL`. |
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| 168. |
What weight of sodium cloride decomposed by `4.9 g` of sulphuric acid, if 6 g of sodium hydrogen sulphide `(NaHSO_(4))` and `1.825 g` of hydrogen chloride are produced in the same reaction ? |
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Answer» The chemical equation for the reaction is : `NaCl+H_(2)SO_(4)rarrNaHSO_(4)+HCl` Weight of the reaction = Weight of NaCl + Weight of `H_(2)SO_(4)` `= x = 4.9 g` (Here `x` is the weight of `NaCl`) Weight of products = Weight of `NaHSO_(4)+` Weight of `HCl` `= 6.0 g+ 1.825 g = 7.825 g` Now, according to law of conservation of mass Weight of reaction = Weight of products `x=4.9=7.825,x=7.825-4.9=2.925g` `:.` Weight of `NaCl` decomposed `= 2.925 g`. |
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| 169. |
Assertion (A): `1 g` of `O_(2)` and `1 g` atom of `O_(3)` have equal number of molecules. Reason (R ): Mass of 1 mol atom is equal to its gram atomic mass.A. If both (A) and (R ) are correct and (R ) is the correct explantion for (A)B. If both (A) and (R ) are correct but (R ) is not the correct explantion for (A)C. If (A) is correct but (R ) is incorrect.D. If (A) and (R ) are incorrect. |
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Answer» Correct Answer - B `1 g` atoms = 1 mol of any compound `= N_(A)` molecules. Both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`. |
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| 170. |
Assetion (A): 1 mol `H_(2)` and `N_(2)` have same volume at same temperature and pressure. Reason (R ): 1 mol gas at `STP` occupies `24.4L` volume.A. If both (A) and (R ) are correct and (R ) is the correct explantion for (A)B. If both (A) and (R ) are correct but (R ) is not the correct explantion for (A)C. If (A) is correct but (R ) is incorrect.D. If (A) and (R ) are incorrect. |
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Answer» Correct Answer - B Both `(A)` and `(R )` are factual statements, but `(R )` is not the correct explanation of `(A)` |
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| 171. |
Iodine can be prepared by the following reactions. `2NaIO_(3) + 5NaSO_(3) rarr 2NaSO_(4) + 2Na_(2) SO_(4) + H_(2)O + I_(2)` How much `NaIO_(3)` is reuired to produce `127 g` is `I_(2)`?A. `1.98 kg`B. `3.96 kg`C. `5.94 kg`D. `0.99 kg` |
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Answer» Correct Answer - A `(Mw "of" I_(2) = 245 g, Mw "of" NaIO_(3) - 198 g, Mw "of" NaHSO_(3) = 104 g)` Mole of `I_(2) = (127)/(254) = 0.5 "mol" I_(2)` `(0.5 "mol" I_(2)) ((2 "mol" NaIO_(3))/("mol" I_(2)))((198 g "of" NaIO_(3))/("mol of" NaIO_(3)))` `implies 0.5 xx 2 xx 198 xx 1980 g = 1.98 kg "of" NaIO_(3)` |
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| 172. |
What mass of NaCl would be decomposed by 98 g of `H_(2)SO_(4)` if 120 g of `NaHSO_(4)` and 27 .5 g of HCl are produced in a reaction. Assuming that law of mass conservation is true :A. 4.95 gB. 49 .5 gC. 0. 495 gD. 495 h |
| Answer» Correct Answer - 2 | |
| 173. |
Iodine can be prepared by the following reactions. `2NaIO_(3) + 5NaSO_(3) rarr 2NaSO_(4) + 2Na_(2) SO_(4) + H_(2)O + I_(2)` How much `NaHSO_(3)` is required to produce `381 g` of `I_(2)`?A. `156.0 g`B. `390.0 g`C. `520.0 g`D. `780.0 g |
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Answer» Correct Answer - D Mol of `I_(2) = (381)/(254) = 1.5 "mol" I_(2)` `(1.5 "mol" I_(2)) ((5 "mol" NaHSO_(3))/("mol" I_(2))) ((104 g NaHSO_(3))/("mol" NaHSO_(3)))` `implies 1.5 xx 5 xx 104 = 780 g "of" NaHSO_(3)` |
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| 174. |
Assertion (A): The equivalent mass of element is varialbe. Reason (R ): It depents on the valency of the element.A. If both (A) and (R ) are correct and (R ) is the correct explantion for (A)B. If both (A) and (R ) are correct but (R ) is not the correct explantion for (A)C. If (A) is correct but (R ) is incorrect.D. If (A) and (R ) are incorrect. |
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Answer» Correct Answer - A Both are factual statements. |
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| 175. |
Assertion (A): Calmel is a chemical compound whereas brass is a mixture. Reason (R ): Calomel always contains 5.6 times as much mercury as chlorine by weight. Brass can be made with widely different ratios of copper and zine.A. If both (A) and (R ) are correct and (R ) is the correct explantion for (A)B. If both (A) and (R ) are correct but (R ) is not the correct explantion for (A)C. If (A) is correct but (R ) is incorrect.D. If (A) and (R ) are incorrect. |
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Answer» Correct Answer - A Calomel is `Hg_(2)Cl_(2)`. (Atomic weight of `Hg` and `Cl` are `= 220.59` and `35.5 g`, respectively) Ratio of weights of `Hg` and `Cl = (2 xx 200.59)/(2 xx 35.5) = 5.6` |
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| 176. |
When phosphours `(P_(4))` is heated in limited amount of `O_(2).P_(4)O_(6)` (tetraphosphorous hexaoxide) is obtained, and in excess of `O_(2)`, `P_(4) O_(10)` (tetraphosphours decaoxide) is obtained. i. `P + 3 O_(2) rarr P_(4) O_(6)`, ii. `P_(4) + 5O_(2) rarr P_(4)O_(10)` What mass of `P_(4) O_(6)` will be produced by the combustion of `2.0 g` of `P_(4)` with `2.0 g` of `O_(2)`.A. 0.0145 molB. 0.0072 molC. 0.029D. 0.0048 |
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Answer» Correct Answer - B Mol of `O_(2) = (2.0)/(31 xx 4) = 0.016 "mol"` This is not enough `O_(2)` to produce `P_(4) O_(10)`, which would have required `5 xx 0.016 = 0.08` mol `O_(2)`. Thus, `(0.016 "mol" P_(4)) ((3 "mol"O_(2))/("mol"P_(4)))` `= 0.048 "mol" O_(2)` used to make `P_(4)O_(6)` Mol of `O_(2)` unreacted `= (0.0625 - 0.048) = 0.0145 "mol" O_(2)` This `O_(2)` can react according to the equation `P_(4) O_(6) + 2 O_(2) rarr P_(4) O_(10` Then, `(0.0145 "mol" O_(2)) ((1 "mol" P_(4) O_(10))/(2 "mol" O_(2))) = 0.00721 "mol" P_(4) O_(10)` the `0.016 "mol of" P_(4) O_(6)` orginally formed, `0.00725` mol is converted to `P_(4) O_(10)`, leaving `(0.016 - 0.00721) = 0.009 "mol" P_(4) O_(6)` `(Mw "of" P_(4) O_(6) = 220)` Weight of `P_(4) O_(6) = 0.009 xx 220 = 1.98 g P_(4) O_(6)` |
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| 177. |
When phosphours `(P_(4))` is heated in limited amount of `O_(2).P_(4)O_(6)` (tetraphosphorous hexaoxide) is obtained, and in excess of `O_(2)`, `P_(4) O_(10)` (tetraphosphours decaoxide) is obtained. i. `P + 3 O_(2) rarr P_(4) O_(6)`, ii. `P_(4) + 5O_(2) rarr P_(4)O_(10)` How many moles of `O_(2)` left unreacted initiallyin reaction (i) ?A. 0.0145 molB. 0.072 molC. 0.029 molD. 0.0048 mol |
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Answer» Correct Answer - A Moles of `O_(2)` left unreacted initially in reaction (i) `= 0.0145 "mol" O_(2)` |
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| 178. |
Each moleof substance A (molar mass = 720) requires 10moles of water for complete hydrolysis and gives B,C and D as the hydrolysed products in a molar ratio of `2:3:2`. If molecular mass of B is 40 and it contributes `40^(@)` of total mass of hydrolysed product then moles pof C obtained will beA. 9B. 13.5C. 3D. 2 |
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Answer» Correct Answer - B |
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| 179. |
5.33 mg of salt `[Cr(H_(2)O)_(5)Cl].Cl_(2)` `H_(2)O` is treated with excess of `AgNO_(3)(aq)` then mass of AgCl ppt. obtained will be : [Given :Cr=52, Cl = 35.5]A. 5.74gmB. 2.87gmC. 4.3gmD. 8.61gm |
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Answer» Correct Answer - A |
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| 180. |
If `0.5 g` of a mixture of two metals. `A` and `B` with respective equivalent weights 12 and 9 displace `560 mL` of `H_(2)` at `STP` from an acid, the composition of the mixture isA. `40% A, 60% B`B. `60% A, 40% B`C. `30% A, 70% B`D. `70% A, 30% B` |
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Answer» Correct Answer - A 1 mol of `H_(2) = 22400 mL = 2 Eq "of" H_(2)` `1 Eq of H_(2) = 11200 mL` Eq of `H_(2) = (560)/(11200) = (1)/(20) Eq` [Let of weight of `A` be weight of `B = 0.5 - x` Eq of `A + Eq "of" B = Eq "of" H_(2)` `(x)/(12) + (0.5)/(9) = (1)/(20)` % of `A = (0.2 xx 100)/(0.5) = 40%` % of `B = 60%` |
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| 181. |
What is the valency of an element of which the eqivalent weight is 12 and the specific heat is 0.25?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B Specific heat `xx` Atomic weight = 6.4 (Dulong and petit law) Atomic weight `= (6.4)/("specific heat") = (6.4)/(0.25)` Atomic weight `= Ew xx` valency = 25.6 Valency `= ("Atomic weight")/(Ew) = (25.6)/(12) = 2` (Valency is always a whole number) |
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| 182. |
10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The molecular formula of the compound is |
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Answer» Correct Answer - `C_(2)H_(6)O` `C_(x)H_(y)O_(z)+(x+(y)/(4)-(z)/(2))O_(2)rarrxCO_(2)+(y)/(2)H_(2)O` Given vol. `10mL" "+" "100mL" " 0" "+0` After reaction `-" "+ 100-10(x+(y)/(4)-(z)/(2))10x-` `100-10(x+(y)/(4)-(z)/(2))+10x=90` `(y)/(4)-(z)/(2)=1` `y-2z=4 " "(1)` Property of `KOH` has absorbed all `CO_(2^(.))` `therefore 10x=20` `x=2` `V.D`. of compound `(C_(x)H_(y)O_(z))=23" "because V.D. = (M_(w))/(2)` `M_(w)=46 " "M_(w)2xx23=46` `12x+y+16z=46` `12xx2+y+16z=46` `y+16z=22" "...(2)` from `(aq). (1) & (2)` `{:(,y-2z=4),(,y+16z=22),(-, "- -"),(,bar(" -18z=-18")):}` `z=1,y=6` Molecular formula `= C_(2)H_(6)O`. |
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| 183. |
25.4g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICI and `ICI_(3)`. Calcualte the number of moles of Icl and `Icl_(3)` formed.A. 0.1 mole, 0.1 moleB. 0.1 mole, 0.2 moleC. 0.5 mole, 0.5 moleD. 0.2 mole, 0.2 mole |
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Answer» Correct Answer - A |
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| 184. |
What weights of `P^(4)O_(6)` and `P_(4)O_(10)` will be produced by the combusion of 31g of `P_(4)` in 32g of oxygen leaving no `P_(4)` and `O_(2)` ?A. 2.75g, 219.5gB. 27.5g, 35.5gC. 55g, 71gD. 17.5g, 190.5g |
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Answer» Correct Answer - B |
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| 185. |
A 200 ml mixture of CO and `CO_2` is passed through excess of red hot charcoal causing the following reaction. `CO_2(g)+C(s) to 2CO(g)` After passing the gas through charcoal , volume increased to 270 ml. Select the correct statement.A. Volume percentage of CO in the original mixture is 65%B. If original mixture was passed through KOH solution volume would have reduced to 70 mlC. Mole fraction of `CO_2` in the original mixture will be 0.25.D. Minimum moles of red hot charcoal required for complete conversion of `CO_2` should be 70 millimoles. |
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Answer» Correct Answer - A::C |
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| 186. |
Four 1-1 litre flasks are seperately filled with the gases `H_(2)`, He and `O_(2)` and `O_(3)` at the same temperature and pressure. The ratio of total number of atomsof these gases present in different flask would be:A. `1:1:1:1`B. `1:2:2:3`C. `2:1:2:3`D. `3:2:2:1` |
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Answer» Correct Answer - C |
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| 187. |
When `CO_(2)(g)` is passed over red hot coke it partially gets reduced to `CO(g)`. Upon passing `0.5` litre of `CO_(2)(g)` over red hot coke, the total volume of the gases increased to `700mL`. The composition of the gaseous mixture at `STP` is :-A. `CO_(2)=200mL:CO=500mL`B. `CO_(2)=350mL:CO=350mL`C. `CO_(2)=0.0mL:CO=700mL`D. `CO_(2)=300mL:CO=400mL` |
| Answer» Correct Answer - D | |
| 188. |
One litre of mixture of `CO` and `CO_(2)` is passed through red hot charcoal in tube. The new volume becomes 1.4 litre. Find out % composition of mixture by volume. All measurements are made at same `P` and `T` |
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Answer» On passing through a charcoal only, `CO_(2)` reduces to `CO`. `CO + C rarr` No reaction volume `a` `CO_(2) + C rarr 2 CO` Volume before reaction `b` `0` volume after reaction 0 `2b` As given `a + b =1` and `a + 2 b = 1.4` `:. b = 0.4 L` `implies % of b = (0.4)/(1) xx 100 = 40%` `:. a = 0.6 L` `implies % of a = (0.6)/(1) xx 100 = 60%` |
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| 189. |
One litre of `CO_(2)` is passed over hot coke. The volume becomes `1.4 L`. Find the composition of products, assuming measurement at `NTP`. |
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Answer» `CO_(2) + C rarr 2CO` `"Initial" 1 L a` `"Final" (1 - a) L 2a` The volume `= (1 - a + 2a) = 1 + a` `:. 1 + a = 1.4` `a = 0.4 L` Volume of `CO_(2)` left `= 1 - 0.4 = 0.6 L` Volume of `CO` formed `= 2 xx 0.4 = 0.8 L` |
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| 190. |
Maximum mass of sucrose `C_12H_22O_11` produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 L `O_2` at 1 atm and 273 K according to given reaction, is : `C(s)+H_2(g)+O_2(g)toC_12H_22O_11(s)`A. 138.5B. 155.5C. 172.5D. 199.5 |
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Answer» Correct Answer - B |
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| 191. |
A crystalline salt on being rendered anhydrous loses 45.6 % of its weight. The percentage composition of anhydrous salt is : `Al = 10.5 % , K = 15.1 % , S = 24.96 %, O = 49.92 %` Find the simplest formula of the anhydrous and crystalline salt. |
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Answer» Correct Answer - Anhydrous salt : `Kal_(2)S_(2)O_(8)` ; Hydrated salt : `KAlS_(2)O_(8).12H_(2)O` Step I. Formula of anhydrous salt `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Al",10.5,27,(10.50)/(27)=0.39,(0.39)/(0.38)=1.03,1),("K",15.1,39,(15.1)/(39)=0.38,(0.38)/(0.38)=1.00,1),("S",24.96,32,(24.96)/(32)=0.78,(0.78)/(0.38)=2.05,2),("O",49.92,16,(49.92)/(16)=3.12,(3.12)/(0.380)=8.21,8):}` Formula of anhydrous salt `= AlKS_(2)O_(8)` Empirical formula mass `= 27 + 39 + 2 xx 32 + 8 xx 16 = 258 u` Step II. Formula of crystalline salt Let the weight of hydrated salt = 100 u , Weight of water lost = 45.6 u Weight of anhydrous salt `= 100 - 45.6 u = 54.4 u` If the weight the anhydrous salt is 54.4 u, that of water = 45.6 u If the weight of anhydrous salt is 258 u, that of water `= ((45.6u))/((54.4u))xx(258u)=216.3u` No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=((216.3u))/((18u))=12` (approx) `:.` The formula of crystalline salt `= KAlS_(2)O_(8).12H_(2)O`. |
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| 192. |
`A` crystalline hydrated sa,t on being rendered anhydrous, loses `45.6%` of its weight. The precentage composition of anhydrous salt is `:Al=10.5%, K=15.1%, S=24.8%` and `I=49.6%`. Find the empirical formula of the anhydrous and crystalline salt : |
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Answer» Correct Answer - `KAIS_(2)O_(8),KAIS_(2)O_(8).12H_(2)O` Empirical formula `KAlS_(2)O_(8){{:(Al,K,S,O,"Elements",),(10.5,15.1,24.8,49.6,"Mass percentage",),(0.388,0.387,0.775,3.1,"Mole ratio",),(1,1,2,8,"Simple ratio",):}` Empirical formula weight `= 258` From weight loss information `:54.4 g` anhydrous salt `-= 45.6 g H_(2)O` ` implies 258g` anhydrous salt `-=216.26 g = 12` mol `H_(2)O` implies Empirical formula of hydrated salt `=KAlS_(2)O_(8).12H_(2)O` |
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| 193. |
Arrange in increasing order of molarity of solute in following solutions consdering water as solvent. (P) 224 gm/L KOH (Q) `11.2%w//v` KOH (R) 5m KOH `(d=0.64gm//ml)`A. `(Q)lt(R)lt(P)`B. `(R)lt(A)lt(P)`C. `(R)lt(P)lt(Q)`D. `(P)lt(Q)lt(R)` |
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Answer» Correct Answer - A |
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| 194. |
Calcualte the volume of `Cl_(2)` gas (in ml) liberated at 1 atm 273 K when 1.74 gm `MnO_(2)` reacts with 2.19 gm HCl according to the following with % yeild 40. `MnO_(2) + 4HCl rightarrow MnCl_(2) + Cl_(2) + 2H_(2)O`A. 336mlB. 112mlC. 134.4mlD. 44.8ml |
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Answer» Correct Answer - C |
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| 195. |
A crystalline compound when heated become anhydrous by losing 51.2 % of the mass. On analysis, the anhydrous compound gave the following percentage composition : `Mg = 20.0 % , S = 26.66 % and O = 53.33 %`. Calculate the molecular formula of the anhydrous compound and crystalline compound. The molecular mass of anhydride compound is 120 u. |
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Answer» Step I. Determination of empirical formula of the anhydrous compound `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Mg",20.0,24,(24.0)/(24)=0.83,(0.83)/(0.83)=1,1),("S",26.66,32,(26.66)/(32)=0.83,(0.83)/(0.83)=1,1),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(0.83)=4.01,4):}` `:.` The empirical formula of anhydrous compound is `MgSO_(4)` Step II. Determination of the molecular formula of the anhydrous compound Empirical formula mass `= 24+32 +4 xx 16 = 24 + 32 + 64 = 120` u Molecular mass = 120 u (Given) , `n=("Molecular mass")/("Empirical formula mass")=(120)/(120)=1` `:.` Molecular formula `=nxx` Empirical formula `=1 xx MgSO_(4)=MgSO_(4)` `:.` Molecular formula of anhydrous compound `= MgSO_(4)` Step III. Determination of molecular formula of the crystalline compound Let us first calculate the number of molecules of water of crystallisation in the compound. For that, Let the weight of hydrated compound = 100 u Weight of water lost = 51.2 u `:.` Weight of anhydrous compound `= 100-51.2 = 48.9` u Now, if the weight of anhydrous compound is 48.8 u, then that of water 51.2 u If the weight of anhydrous compound is 120 u, then that of water `= (51.2)/(48.8)xx120 = 125.9` u Molecular mass of water `(H_(2)O)=2xx1+16 = 18` u `:.` No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=(125.9)/(18)=7` (approx) Thus, the formula of crystalline compound `= MgSO_(4).7H_(2)O`. |
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| 196. |
An organic compound has the following percentage composition , `C = 48 %, H = 8 %, N = 28 %`. Calculate the empirical formula of the compound. |
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Answer» For the available data : Percentage of `C = 48%` , Percentage of `H = 8 %` , Percentage of `N = 28 %` Total percentage of C, H and `N = 48 + 8 + 28 = 84 %` We know that the sum of the percentages must be 100. But in this case, it is only `84 %`. The balance `(100-84)= 16 %` is regarded as the percentage of `"oxygen"^(**)` in the compound Step I. Calculation of simplest whole number ratios of the elements `{:("Element","Percentage","Atomic Mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",48.0,12,(47.0)/(12)=4,(4)/(1)=4,4),("H",8.0,1,(8.0)/(1)=8,(8)/(1)=8,8),("N",28.0,14,(28.0)/(14)=2,(2)/(1)=2,2),("O",16.0,16,(16.0)/(16)=1,(1)/(1)=1,1):}` The simplest whole number ratios of different elements are : `C:H:N:O : : 4 : 8 : 2 : 1` Step II. Writing the empirical formula of the compound The empirical formula of compound `= C_(4)H_(8)N_(2)O`. |
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| 197. |
`10 mL` of `0.2 N HCl` and `30 mL` of `0.1 N HCl` to gether exaclty neutralises `40 mL` of solution of `NaOH`, which is also exactly neutralised by a solution in water of `0.61 g` of an organic acid. What is the equivalent weight of the organic acid?A. 61B. 91.5C. 122D. 183 |
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Answer» Correct Answer - C `10 mL` of `0.2 N HCl + 30 mL` of `0.1 N HCl -= 40 mL` of `NaOH (-= 0.61 g` organic acid) mEq of `HCl -= mEq` of `NaOH -= mEq` of orgainc acid `10 xx 0.2 + 30 xx 0.1 -= (0.61)/(e) xx 100` `5 = (0.61 xx 1000)/(E)` `E = (610)/(5) = 122` |
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| 198. |
0.9 gm of a volatile solid organic compound (molecular weight =90) containing carbon, hydrogen and oxygen was heated with 224 ml of oxygen at 1 atm and `0^(@)` c. After combustion, the total volume of gases was 560 ml at same T and P. On treatment with KOH, the volume decreased to 112ml. Determine the value of x+y+z if molecular formula of organic compound is `C_(x)H_(y)O_(z).` |
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Answer» Correct Answer - 8 |
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| 199. |
A gaseous hydrocarbon `X`, was burnt in excess of oygen. A `0.112 dm^(2)` sample of `X`, at `STP` gave `0.88 g` of `CO_(2)`. How many C-atoms are there in one molecule of `X`? a. 1 b. 2 c. 3 d. 4 |
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Answer» d. `C_(x) H_(y) (g) + (x + (y)/(4)) O_(2) (g) rarr xCO_(2) (g) + (y)/(2) H_(2) O(l)` M moles `= (0.112)/(22.4) xx 10^(3) = 5` `= 5 x` some left `O_(2)` `5x` - `:. 5 x = (0.88 xx 1000)/(44) = 20` `:. x = 4` |
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| 200. |
44 g of a sample of organic compound on complet combustion gives 88 g `CO_(2)` and 36 g of `H_(2)O` . The molecular formula of the compound may be :-A. `C_(4)H_(6)`B. `C_(2)H_(6)O`C. `C_(2)H_(4)O`D. `C_(3)H_(6)O` |
| Answer» Correct Answer - 3 | |