Question

the boiling point of a solution of 0.11 gram of a substance in 15 gram of ether was found to be 0.1 degree Celsius higher than that of pure Ether the molecular weight of the substance will be​

Answer 1

Answer : The molecular weight of the substance will be​, 148.13 g/mole

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$\Delta T_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}$

where,

$\Delta T_b$ = change in boiling point = $0.1^oC$

$k_b$ = boiling point constant  = $2.02^oCKg/mole$

m = molality

$w_2$ = mass of solute = 0.11 g

$w_1$ = mass of solvent (ether) = 15 g

$M_2$ = molar mass of solute = ?

Now put all the given values in the above formula, we get the molecular weight of the substance.

$0.1^oC=\frac{1000\times 2.02^oCKg/mole\times 0.11g}{15g\times M_2}$

$M_2=148.13g/mole$

Therefore, the molecular weight of the substance will be​, 148.13 g/mole