Question

Calculate the weight of non-volatile solute having molecular weight 40, which should be dissolved in $57gm$ octane to reduce its vapour pressure to $80\%$.

• A. $47.2gm$
• B. $4gm$
• C. $106.2gm$
• D. none of these

Answer 1

Answer: (B)

$4gm$

Let the vapour pressure of pure octane be $p_1^0$.

Then, the vapour pressure of the octane after dissolving the non-volatile solute $=80/100$ $p_1^0=$ 0.8 $p_1^0$.

Molar mass of solute, $M_2=$ 40 g/mol

Mass of octane, $w_1=$ 114 g

Molar mass of octane, ($C_8{H}_{18}$), $M_1$ $=8\times 12+18\times 1=114$ g/mol

Applying the relation,

$\frac{p_1^0-p_1}{p_1^0}$ $=\frac{w_2\times M_1}{M_2\times w_1}$

$\frac{p_1^0-0.8p_1^o}{p_1^0}$ $=\frac{w_2\times 114}{40\times 57}$

⇒ $0.2p_1^0/p_1^0=w_2/20$

⇒ $0.2=w_2/20$

⇒ $w_2=4g$

Hence, the required mass of the solute is 4g.