1.

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)]

Answer»

`5.8 g mol^(-1)`
`0.58 g mol^(-1)`
`58 g mol^(-1)`
`0.88 g mol^(-1)`

Solution :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)`


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