`1.4 g` of acetone dissolved in `100 g` of benzene gave a solution which freezes at `277.12 K`. Pure benzene freezes at `278.4 K`.`2.8` of solid `(A)` dissolved in `100 g` of benzene gave a solution which froze at `277.76 K`. Calculate the molecular mass of `(A)`.

Answer

232

Explanation

For acetone +benzene mixture
`DeltaT=(K._fxx1000 xxw)/(mxxW)`
`(278.277.12)=(1000 xxK_f xx1.4)/(100xx58)`
`1.28 =(1000 xxK_f xx1.4)/(100 xx58)" "....(i)`
For solute (A)+benzene mixture
Suppose m is the molecular weight of (A).
(278.40 - 277.76) `=(1000 xx mK_fxx 2.8)/(100 xx m)`
`0.64 =( 1000 xxK_fxx 2.8)/(100 xxm) " "... (ii) `
From equation (i) and (ii), we get m= 232.