Objective question (single correct answer). i. The molarity of a aqueous solution of glucose `(C_(6) H_(12) O_(6))` is 0.01 To `200 mL` of the solution, which of the following should be carried out to make it `0.02 M`? I. Evaporate `50 mL` of solution III. Add `0.180 g` of glucose and then evaporate `50 mL` of solution III. Add `50 mL` of water The correct option is: a. I b. III c. II d. I, II, III ii. The atomic mass of `Cu` is 63.546. There are only two naturally occuring isotopes of copper `Cu^(63)` and `Cu^(65)`. The percentage of natural abundance of `Cu^(63)` in nearly a. 30 b. 10 c. 50 d. 73 iii. An aqueous solution of urea `(NH_(2) COHN_(2))` is 3.0 molal. The mole fraction of urea is a. 0.33 b. 0.25 c. 0.66 d. 0.05 iv `0.2 M H_(2) SO_(4)` `(1 mL)` is diluted to 1000 times of its initial volume. the final normality of `H_(2) SO_(4)` is: a. `2 xx 10^(-3)` b. `2 xx 10^(-4)` c. `4 xx 10^(-4)` d. `2 xx 10^(-2)` v. Which of the following question are dependant on temperature? a. Molarlity b. Normality c. Mole fraction d. Molality vi. A sample of `H_(2) SO_(4)` density `1.85 mL^(-1)` is 90% by weight. What is the volume of the acid that has to be used to make `1 L` f `0.2 M H_(2) SO_(4)`? a. `16 mL` b. `18 mL` c. `12 mL` d. `10 mL` vii. The hydrated salt `Na_(2)SO_(4)`. `nH_(2)O` undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of `n` will be a. 5 b. 7 c. 3 d. 10 viii. 0.2 mol of `HCl` and 0.1 mol of barium chloride is dissolved in water to produce a `500 mL` solution. The molarity of `Cl^(ɵ)` is. a. `0.06 M` b. `0.12 M` c. `0.09 M` d. `0.80 M` ix. The density of `1 M` solution of `NaCl` is `1.055 g mL^(-1)`. The molality of the solutions is. a. 1.0585 b. 1.00 c. 0.0585 d. 0.10 x. Hydrochloric acid solution `A` and `B` have concentration of `0.5 N`, and `0.1 N`, respectively. The volume of solutions `A` and `B` required to make `2 L` of `0.2 N` hydrochloric acid are a. `0.5 L of A + 1.5 of B` b. `1.0 L of A + 1.0 L of B` c. `0.75 L of A + 1.25 L of B` d. `1.5 L of A + 0.5 L of B`

Objective question (single correct answer). i. The molarity of a aqueo

1.	Objective question (single correct answer). i. The molarity of a aqueous solution of glucose `(C_(6) H_(12) O_(6))` is 0.01 To `200 mL` of the solution, which of the following should be carried out to make it `0.02 M`? I. Evaporate `50 mL` of solution III. Add `0.180 g` of glucose and then evaporate `50 mL` of solution III. Add `50 mL` of water The correct option is: a. I b. III c. II d. I, II, III ii. The atomic mass of `Cu` is 63.546. There are only two naturally occuring isotopes of copper `Cu^(63)` and `Cu^(65)`. The percentage of natural abundance of `Cu^(63)` in nearly a. 30 b. 10 c. 50 d. 73 iii. An aqueous solution of urea `(NH_(2) COHN_(2))` is 3.0 molal. The mole fraction of urea is a. 0.33 b. 0.25 c. 0.66 d. 0.05 iv `0.2 M H_(2) SO_(4)` `(1 mL)` is diluted to 1000 times of its initial volume. the final normality of `H_(2) SO_(4)` is: a. `2 xx 10^(-3)` b. `2 xx 10^(-4)` c. `4 xx 10^(-4)` d. `2 xx 10^(-2)` v. Which of the following question are dependant on temperature? a. Molarlity b. Normality c. Mole fraction d. Molality vi. A sample of `H_(2) SO_(4)` density `1.85 mL^(-1)` is 90% by weight. What is the volume of the acid that has to be used to make `1 L` f `0.2 M H_(2) SO_(4)`? a. `16 mL` b. `18 mL` c. `12 mL` d. `10 mL` vii. The hydrated salt `Na_(2)SO_(4)`. `nH_(2)O` undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of `n` will be a. 5 b. 7 c. 3 d. 10 viii. 0.2 mol of `HCl` and 0.1 mol of barium chloride is dissolved in water to produce a `500 mL` solution. The molarity of `Cl^(ɵ)` is. a. `0.06 M` b. `0.12 M` c. `0.09 M` d. `0.80 M` ix. The density of `1 M` solution of `NaCl` is `1.055 g mL^(-1)`. The molality of the solutions is. a. 1.0585 b. 1.00 c. 0.0585 d. 0.10 x. Hydrochloric acid solution `A` and `B` have concentration of `0.5 N`, and `0.1 N`, respectively. The volume of solutions `A` and `B` required to make `2 L` of `0.2 N` hydrochloric acid are a. `0.5 L of A + 1.5 of B` b. `1.0 L of A + 1.0 L of B` c. `0.75 L of A + 1.25 L of B` d. `1.5 L of A + 0.5 L of B`
Answer» Correct Answer - A::B::C::D i. c. Since the molarity of glucose has to be increased (from `0.01 M` to `0.02 M`), so this can be carried out either by evaporating the solution or by adding some more glucose. So, the only possible answers is (c ). I. Evaporate `50 mL` of solution mmoles of glucose initially `= 0.01 xx 200 = 2` volume after evaporation `= 200 - 50` `= 150 mL` `M_("glucose") = (("mmoles")/(V_(mL))) = (2)/(150) = 0.013 M` II. mmoles of glucose added `= (0.180)/(180) xx 100 = 1` Total mmoles of glucose `= 2 + 1 = 3` Volumes `= 150 mL` (after evaporation of `50 mL` solution) `M_("glucose") = (3)/(150) = 0.02 M` III. Add `50 mL` of water New volume of solution `= 200 + 50 = 250 mL` `M_("glucose") = (2 mmol)/(250 mL) = 0.008 M` Hence, answer is (c ) ii. d. `63.546 = (a xx 63 + (100 - a) xx 65)/(100)` `a = 72.7% ~~ 73.%` iii. d. `m = (X_(2) xx 1000)/((1 - x_(2)) xx Mw_(1))` `3 = (X_(2) xx 1000)/((1 - x_(2)) xx 18)` Solve for `x_(2) = 0.05` iv. c. `M_(1) V_(1) = M_(2) V_(2)` `0.2 M xx 1 mL = M_(2) xx 1000 mL` `M_(2) = 2 xx 10^(-4)` `:. N = 2 xx 2 xx 10^(-4) = 4 xx 10^(-4)` vi. c. `M_(1) V_(1) = M_(1) V_(2) (M = (% "by weight" xx 10 xx d)/(Mw_(2)))` `V_(1) xx (90 xx 10 xx 1.8)/(98) = 0.2 xx 1 L` `V_(1) = 0.012 L = 12 mL` vii. d. Loss in weight is due to `nH_(2) O`. `:. (142 + 18 n) g of Na_(2) SO_(4). nH_(2) O = 18 n g` of loss in weight of `H_(2) O` `100 g of Na_(2) SO_(4) . nH_(2) O = (18 n xx 100)/(142 + 18 n)` `:. (18 n xx 100)/(142 + 18n) = 55.9` solve for `n implies n = 9.99 ~~ 10` viii. d. `0.02 "mol" HCl = 0.02 of H^(o+) + 0.2 "mol" Cl^(ɵ)` `0.01 "mol" BaCl_(2) = 0.1 "mol" of Ba^(2+) + 0.1 xx 2 "mol" of Cl^(ɵ)` Total `Cl^(ɵ) = 0.4 "mol"`. Total volume `= 500 mL = (1)/(2) L` `:. [Cl^(ɵ)] = (0.4)/(1//2) = 0.8 M` ix. b. `d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))` `1.0585 = 1 M ((58.5)/(1000) + (1)/(m))` solve for `m` `m = 1.0` x. a. `V_(1) + V_(2) = 2 L` `N_(1) V_(1) + N_(2) V_(2) = N_(3) V_(3)` `0.5 xx V_(1) + 0.1 xx V_(2) = 0.2 xx 2` `0.5 V_(2) + 0.1 V_(2) = 0.4` Solve equations (i) and (ii) `V_(1) = 0.5 L, V_(2) = 1.5 L`

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