1
MCQ (Single Correct Answer)

NEET 2016 Phase 1

At 100oC the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be
A
102oC
B
103oC
C
101oC
D
100oC

Explanation

Given that

ws = 6.5 g, wA = 100 g

ps = 732 mm of Hg

kb = 0.52, Tob = 100oC

po = 760 mm of Hg

$${{{p^o} - {p_s}} \over {{p^o}}} = {{{n_2}} \over {{n_1}}}$$

$$ \Rightarrow $$ $${{760 - 732} \over {760}} = {{{n_2}} \over {{{100} \over {18}}}}$$

$$ \Rightarrow $$ n2 = 0.2046 mol

$$\Delta $$Tb = Kb × m

Tb - Tob = $${k_b} \times {{{n_2} \times 1000} \over {{w_{A\left( g \right)}}}}$$

$$ \Rightarrow $$ Tb - 100oC = $${{0.52 \times 0.2046 \times 1000} \over {100}}$$ = 1.06

$$ \Rightarrow $$ Tb = 101.06 oC
2
MCQ (Single Correct Answer)

NEET 2016 Phase 2

The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
A
0
B
1
C
2
D
3

Explanation

Ba(OH)2 is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.

Ba(OH)2(aq) $$ \to $$ Ba2+(aq) + 2OH(aq)

Thus, van’t Hoff factor i = 3.
3
MCQ (Single Correct Answer)

NEET 2016 Phase 2

Which one of the following is incorrect for ideal solution?
A
$$\Delta {H_{mix}} = 0$$
B
$$\Delta {U_{mix}} = 0$$
C
$$\Delta P = {P_{obs}} - P$$calculated by Raoult's law
D
$$\Delta {G_{mix}} = 0$$

Explanation

For ideal solution, we have

$$\Delta $$Hmix = 0, $$\Delta $$Vmix = 0

Now Umix = ∆Hmix – P$$\Delta $$Vmix

$$ \therefore $$ $$\Delta $$Umix = 0

Also, for an ideal solution,

pA = xApAo, pB = xBpBo

$$ \therefore $$ $$\Delta $$p = pobserved – pcalculated = 0

$$\Delta $$Gmix = $$\Delta $$Hmix – T$$\Delta $$Smix

For an ideal solution, $$\Delta $$Smix $$ \ne $$ 0

$$ \therefore $$ $$\Delta $$Gmix $$ \ne $$ 0
4
MCQ (Single Correct Answer)

AIPMT 2015 Cancelled Paper

Which one of the following electrolytes has the same value of van't Hoff factor (i) as that of Al2(SO4)3 (if all are 100% ionised)?
A
Al(NO3)3
B
K4[Fe(CN)6]
C
K2SO4
D
K3[Fe(CN)6]

Explanation

Ai2(SO4)3 ⇌ 2Al+3 + 3SO42–

van't Hoff factor, i = 5

K2SO4 ⇌ 2K+ + SO42–

van't Hoff factor, i = 3

K3[Fe(CN)6] ⇌ 3K+ + [Fe(CN)6]2-

van't Hoff factor, i = 4

Al(NO3)3 ⇌ Al3+ + 3NO3

van't Hoff factor, i = 4

K4[Fe(CN)6] ⇌ 4K+ + [Fe(CN)6]4–

van't Hoff factor, i = 5

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