1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg moll. The molar mass of the solute is

Options

A. 256 kg/mol
B. 256 g mol
C. 256 g/mol
D. 256 mg/mol

Answer

C

Explanation

We have `M _(2) = ( K _(f) xx W _(2) xx 1000)/( Delta T _(f) xx W _(1))`
when `W _(1) =` weight of solvent = 50 g
`W _(2)` weight of olute `= 1. 00 g`
`K _(f) = ` molar depression constant
`= 5. 12 K kg mol ^(-1)`
Subsituting the values of various terms involved in the above equation, we get
`M _(2) = ( 5 12 K kg mol ^(-1) xx 1. 00 g xx 1000 g kg ^(-1))/( -0. 40 xx 50 g)= 256 g mol ^(-1)`
Thus, molar mass of the solute = 256 g/mol