If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80 %, the molar mass (in g mol-1) of the solute is [Given that molar mass of n-octane is 114 g mol-1]

Question

If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80 %, the molar mass (in g mol-1) of the solute is [Given that molar mass of n-octane is 114 g mol-1] (A) 40 (B) 60 (C) 80 (D) 20

Tardigrade Admin · Accepted Answer

Assuming dilute solution, (P0-Ps/Ps) ≈(P0-Ps/P0)=(n textSolute/n textSolvent) Let P0=100, V.P. reduced to 80 %, ∴ Ps=80 (100-80/100)=(8 m/114 114) m=40

Q. If $8 g$ of a non-electrolyte solute is dissolved in $114 g$ of n-octane to reduce its vapour pressure to $80%$ , the molar mass (in $g m o l^{- 1})$ of the solute is [Given that molar mass of n-octane is $114 g m o l^{- 1}]$

$40$

$60$

$80$

$20$

Assuming dilute solution,
$\frac{P _{0} - P _{s}}{P _{s}} \approx \frac{P _{0} - P _{s}}{P _{0}} = \frac{n _{Solute}}{n _{Solvent}}$
Let $P_{0} = 100, V . P .$ reduced to $80%$ ,
$∴ P_{s} = 80$
$\frac{100 - 80}{100} = \frac{8 m}{114114}$
$m = 40$

Q. If 8g of a non-electrolyte solute is dissolved in 114g of n-octane to reduce its vapour pressure to 80%, the molar mass (in gmol−1) of the solute is [Given that molar mass of n-octane is 114gmol−1]

40

60

80

20