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1

NEET 2016 Phase 2

MCQ (Single Correct Answer)
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.
2

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
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
3

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
Which of them is not equal to zero for an ideal solution?
A
$$\Delta {V_{mix}}$$
B
$$\Delta P = {P_{observed}} - {P_{Raoult}}$$
C
$$\Delta {H_{mix}}$$
D
$$\Delta {S_{mix}}$$

Explanation

For an ideal solution, $$\Delta $$Smix > 0 while $$\Delta $$Hmix, $$\Delta $$Vmix and $$\Delta $$P all are 0.
4

AIPMT 2015

MCQ (Single Correct Answer)
What is the mole fraction of the solute in a 1.00 m aqueous solution?
A
1.770
B
0.0354
C
0.0177
D
0.177

Explanation

Number of moles of water in 1000 g = $${{1000} \over {18}}$$ = 55.56 mol

Thus, mole fraction of solute = $${1 \over {1 + 55.56}}$$ = 0.0177

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