The boiling a point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. `K_(b)` for benzene is 2.53 K kg `mol^(-1)`.

Options

A. `5.8 g mol^(-1)`
B. `0.58 g mol^(-1)`
C. `58 g mol^(-1)`
D. `0.88 g mol^(-1)`

Answer

C

Explanation

The elevation `(Delta T_(b))` in the boiling point `= 354.11 K - 353.23 K = 0.88 K`
Substituting these values in expression
`M_("Solute")=(K_(b)xx1000xxw)/(Delta T_(b)xx W)`
Where, w = weight of solute, W = weight of solvent
`M_("solute")=(2.53xx1.8xx1000)/(0.88xx90)=58 gm mol^(-1)`
Hence, molar mass of the solute `= 58 gm mol^(-1)`