Question

When 10g of a non-volatile solute is dissolved in 100g of benzene, it raises the boiling point by \[{\rm{1^\circ C}}\] then the molecular mass of the solute is (\[{{\rm{k}}_{\rm{b}}}\] for benzene = \[{\rm{2}}{\rm{.53^\circ Ckgmol}}{{\rm{l}}^{{\rm{ - 1}}}}\])
A. 223g
B. 233g
C. 243g
D. 253g

Answer 1

Hint: When a non-volatile solute is added to a solvent, the boiling point of this solution increases. This is depicted as elevation in boiling point. It only depends on the number of solute particles, not on their nature.

Formula Used:
\[\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{b}}}{\rm{ \times 1000 \times }}{{\rm{w}}_{\rm{2}}}}}{{{{\rm{M}}_{\rm{2}}}{\rm{ \times }}{{\rm{w}}_{\rm{1}}}}}\]
where m=molality.
\[{{\rm{k}}_{\rm{b}}}\]=Boiling Point Elevation Constant or Molal Elevation Constant or Ebullioscopic Constant
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\]=elevation of the boiling point
\[{{\rm{M}}_{\rm{2}}}\]=molecular mass of the non-volatile solute
\[{{\rm{w}}_{\rm{2}}}\]=mass of solute
\[{{\rm{w}}_{\rm{1}}}\]=mass of solvent

Complete Step by Step Solution:
In this question, we are given the masses of non-volatile solute and solvent which is benzene, elevation in boiling point, and the ebullioscopic constant.

Given,
Mass of solute=10 g=\[{{\rm{w}}_{\rm{2}}}\]
Mass of solvent=100 g=\[{{\rm{w}}_{\rm{1}}}\]
We have to find the molecular mass of the non-volatile solute.
\[{{\rm{M}}_{\rm{2}}}\]=molecular mass of the non-volatile solute
\[{{\rm{k}}_{\rm{b}}}\]=Molal Elevation Constant or Ebullioscopic constant=\[{\rm{2}}{\rm{.53^\circ CKgmol}}{{\rm{l}}^{{\rm{ - 1}}}}\]

Molal elevation constant is the increase in boiling point due to the dissolution of one mole of a non-volatile solute in 1 kg or 1000g of a solvent.
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\]=elevation of the boiling point =1°C
\[\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{b}}}{\rm{ \times 1000 \times }}{{\rm{w}}_{\rm{2}}}}}{{{{\rm{M}}_{\rm{2}}}{\rm{ \times }}{{\rm{w}}_{\rm{1}}}}}\]

For \[{{\rm{M}}_{\rm{2}}}\], the above expression can be rewritten as
\[{{\rm{M}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{b}}}{\rm{ \times 1000 \times }}{{\rm{w}}_{\rm{2}}}}}{{{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ \times }}{{\rm{w}}_{\rm{1}}}}}\]
=\[\frac{{{\rm{2}}{\rm{.53^\circ CKg/mol \times 1000g/kg \times 10g}}}}{{{\rm{1^\circ C \times 100g}}}}\]
= 253g/mol
So, option D is correct.

Note: While attending to the question, one must note the unit of the molal elevation constant. In this question, the molar elevation constant is in \[{\rm{^\circ Ckgmol}}{{\rm{l}}^{{\rm{ - 1}}}}\] . So, it is not required to convert the elevation in boiling point into Kelvin. But in case the ebullioscopic constant was given in K Kg/mol, it will be necessary to convert the given temperature into K. The units of the given quantities must be mentioned properly.