On mixing , heptate and octane form an ideal solution. At 373 K. The vapour pressure of the two liquid components (heptaane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtainde by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane=100 g `mol^(-1)` )

Options

A. 0.0186 K
B. 0.0372 K
C. 0.0558 K
D. 0.0744 K

Answer

c

Explanation

Fro `NaSO4, i=3`
Since `alpha` is eaual to 1
`"Molality (m)"=((0.01mol))/(1kg)`
`DeltaT_(f)=ixxK_(f)xxm`
`=3xx(1.86 " K kg"mol^(-1)) xx (0.01 mol kg^(-1))`
=0.0558 K.