Question

At $293K$, ethyl acetate has vapour pressure of $72.8$ torr of $Hg$ and ethyl propionate has vapour pressure of $27.7$ torr of $Hg$. Assuming their mixtures to obey Raoult's law, determine the vapour pressure of a mixture containing $25g$ of ethyl acetate and $50g$ of ethyl propionate.

• A. 44.24
• B. 38.56
• C. 57.28
• D. 62.84

Answer 1

The correct option is A 44.24

Moles of ethyl acetate= $n_A=\frac{25}{88}=0.284$

Moles of ethyl propionate= $n_B=\frac{50}{102}=0.490$

Mole fraction of ethyl acetate= $x_A=\frac{0.284}{0.284+0.490}=0.367$

Mole fraction of ethyl propionate= $x_B=\frac{0.490}{0.774}=0.633$

$P=x_A{P}_A^{\cdot }+x_B{P}_B^{\cdot }=0.367\times 72.8+0.633\times 27.7=44.24torr$ of $Hg$