A sample of 20.0 g of a compound (molecular weight 120) which is a non-electrolyte is dissolved in 10.0 grams of ethanol (C2H5OH) If the vapour pressure of pure ethanol at the temperature used is 0.250 atm, what is the vapour pressure of ethanol above the solution?

Question

A sample of 20.0 g of a compound (molecular weight 120) which is a non-electrolyte is dissolved in 10.0 grams of ethanol (C2H5OH) If the vapour pressure of pure ethanol at the temperature used is 0.250 atm, what is the vapour pressure of ethanol above the solution? (A) 0.250 atm (B) 0.83 atm (C) 0.125 atm (D) 0.141 atm

Tardigrade Admin · Accepted Answer

(Po-Ps/Po)=(n/n+N) (0.25-Ps/0.25)=((20/120)/(20)120+(10)46=(0.167/0.167+0.217) ∴ 0.25-P/s)=(0.25×0.167/0.167+0.217) ∴ Ps=0.141

Q. A sample of 20.0 g of a compound (molecular weight 120) which is a non-electrolyte is dissolved in 10.0 grams of ethanol $(C_{2} H_{5} O H)$ If the vapour pressure of pure ethanol at the temperature used is 0.250 atm, what is the vapour pressure of ethanol above the solution?

0.250 atm

0.83 atm

0.125 atm

0.141 atm

$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{n}{n + N}$
$\frac{0.25 - P _{s}}{0.25} = \frac{\frac{20}{120}}{\frac{20}{120} + \frac{10}{46}} = \frac{0.167}{0.167 + 0.217}$
$∴ 0.25 - P_{s} = \frac{0.25 \times 0.167}{0.167 + 0.217}$
$∴ P_{s} = 0.141$

Q. A sample of 20.0 g of a compound (molecular weight 120) which is a non-electrolyte is dissolved in 10.0 grams of ethanol (C2​H5​OH) If the vapour pressure of pure ethanol at the temperature used is 0.250 atm, what is the vapour pressure of ethanol above the solution?