Question

At 293 K, methanol has a vapor pressure of 97.7 Torr and ethanol has a vapor pressure of 44.6 Torr. What would be the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K?

Answer 1

Answer: The vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction in solution  

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$$p_{total}=x_Ap_A^0+x_BP_B^0$

moles of ethanol=$\frac{\text{Given mass}}{\text {Molar mass}}=\frac{80g}{46g/mol}=1.7moles$

moles of methanol= $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{97g}{32g/mol}=3.0moles$

Total moles = moles of ethanol + moles of methanol = 1.7 +3.0 = 4.7

$x_{ethanol}=\frac{1.7}{4.7}=0.36,$

$x_{methanol}=1-x_{ethanol}=1-0.36=0.64$

$p_{ethanol}^0=44.6torr$

$p_{methanol}^0=97.7torr$

$p_{total}=0.36\times 44.6+0.64\times 97.7=78.3torr$

Thus the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.