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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. Ib. IIIc. IId. 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. 30b. 10c. 50d. 73 iii. An aqueous solution of urea (NH_(2) COHN_(2)) is 3.0 molal. The mole fraction of urea is a. 0.33b. 0.25c. 0.66d. 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. Molarlityb. Normalityc. Mole fractiond. 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 mLb. 18 mLc. 12 mLd. 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. 5b. 7c. 3d. 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 Mb. 0.12 Mc. 0.09 Md. 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. 1.0585b. 1.00c. 0.0585d. 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 |
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Answer» 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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