Methanol `(p^@"=90 mm Hg at 300 K")` and ethanol `(p^@="51 mm Hg at 300 K")` form very nearly an ideal solution. The total vapour pressure of a solution obtained by mixing 23 g ethanol with 32g methanol would be

Options

A. 17mmHg
B. 60mmHg
C. 77mmHg
D. 82mmHg

Answer

C

Explanation

Amount of methanol, `n_(1)= (m_(1))/(M_(1))= (32g)/(32 g mol^(-1))= 1.0 mol`
Amount of ethanol, `n_(2)= (m_(2))/(M_(2))=(23g)/(46 g mol^(-1))= 0.5mol`
Mole fraction of methanol, `x_(1)= (n_(1))/(n_(1) + n_(2))= (1.0 mol)/((1.0+0.5) mol)= 2//3`
Total vapour pressure, `p= x_(1)p_(1) +x_(2)p_(2)= [(2//3) (90)+ (1//3) (51)]` mg Hg= 77 mmHg