0.48g of a substance was dissolved in 10.6 g `C_6H_6.` The freezing point of benzene was lowered by `1.8^@C.` Calculate molar mass of the substance. Molecular depression constant for benzene is `"5K kg mol"^(-1)`.

Options

A. `153.6g mol^(-1)`
B. `125.79g mol^(-1)`
C. `251.6g mol^(-1)`
D. `89.42g mol^(-1)`

Answer

B

Explanation

`because Delta T= (100K_(f)w)/(mW)`
`because w=0.48g, W=10.6g, Delta T_(f)= 1.8, K_(f)= 5K kg mol^(-1)`
`therefore 1.8= (1000 xx 5 xx 0.48)/(m xx 10.6) therefore m=125.79 g mol^(-1)`