Question

Calculate the mass of a non-volatile solute (molar mass $40 g mol^{–1}$) which should be dissolved in 114 g octane to reduce its vapour pressure to $80\%$

Answer 1

The correct answer is: $8 g$
Let the vapour pressure of pure octane be $p^o_1.$
Then, the vapour pressure of the octane after dissolving the non-volatile solute is $\frac{80}{100} p^o_1 = 0.8p^o_1.$
Molar mass of solute, $M_2 = 40 g mol^{- 1}$
Mass of octane, $w_1 = 114 g$
Molar mass of octane, $(C_8H_{18}), M_1 = 8 × 12 + 18 × 1 = 114 g mol^{- 1}$
Applying the relation,
$\frac{ (p^o_1-p_1)}{ p^o_1}= \frac{(w_2\times M_1)}{(M_2 \times w_1)}$
$⇒ \frac{(p^o_1-0.8p_1)}{ p^o_1}= \frac{(w_2\times 114)}{(40 \times 114)}$
$⇒\frac{0.2p^o_1}{p^o_1} = \frac{w_2}{40}$
$⇒0.2= \frac{w_2}{40}$
$⇒w_2 = 8g$
Hence, the required mass of the solute is $8 g$.