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

A, B, C, D

Explanation

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`