Respuesta :
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.
[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]
where, x = mole fraction in solution
[tex]p^0[/tex] = pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2[/tex][tex]p_{total}=x_Ap_A^0+x_BP_B^0[/tex]
moles of ethanol=[tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{80g}{46g/mol}=1.7moles[/tex]
moles of methanol= [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{97g}{32g/mol}=3.0moles[/tex]
Total moles = moles of ethanol + moles of methanol = 1.7 +3.0 = 4.7
[tex]x_{ethanol}=\frac{1.7}{4.7}=0.36,[/tex]
[tex]x_{methanol}=1-x_{ethanol}=1-0.36=0.64[/tex]
[tex]p_{ethanol}^0=44.6torr[/tex]
[tex]p_{methanol}^0=97.7torr[/tex]
[tex]p_{total}=0.36\times 44.6+0.64\times 97.7=78.3torr[/tex]
Thus the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.
