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1

### 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.
2

### 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
3

### AIPMT 2015 Cancelled Paper

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

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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Class 12
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