Question

The boiling point of benzene is 353.23 K. When 1.80 gm of a nonvolatile silute was dissolved in 90 gm of benzene, the boiling point is raised to 354.11 K. The molar mass of the solute is [`K_(b)` for benzene `= 2.53 K mol^(-1)`]
A. `5.8 g mol^(-1)`
B. `0.58 g mol^(-1)`
C. `58 g mol^(-1)`
D. `0.88 g mol^(-1)`

Answer 1

Correct Answer - C
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)`