Question

Relative decrease in vapor pressure is 0.4 for a solution containing 1 mole AB in 3 mol $H_{2}O$, therefore AB is ionised at the extent(percentage) of:

• A. $60$
• B. $50$
• C. $100$
• D. $40$

Answer 1

The correct option is A $60$

we have,

relative lowering in vapour pressure = $x_A$ = $\frac{1}{1+3}$

0.4 = i * 0.25

where i = Vont Hoff factor

hence i = 1.6

and we have,

i = 1 + $\alpha (n-1)$

here n = 2

we have,

1.6 = 1 + $\alpha \ast 1$

so $\alpha$ = 0.6 or 60%