Question

Calculate the boiling point of a solution containing 1.04g glucose (molar mass
= 180g mol-') lies in 80.2g water. [Boiling point of pure water = 373.15K, K,
for water = 0.52K kg mol-']
​

Answer 1

Answer : The boiling point of a solution is, 373.18 K

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$T_b-T^o_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of pure water = 373.15 K

$k_b$ = boiling point constant  = 0.52 K kg/mole

m = molality

$w_2$ = mass of solute (glucose) = 1.04 g

$w_1$ = mass of solvent (water) = 80.2 g

$M_2$ = molar mass of solute (glucose) = 180 g/mole

Now put all the given values in the above formula, we get the boiling point of a solution.

$T_b-373.15K=\frac{1000\times 0.52Kkg/mole\times 1.04g}{80.2g\times 180g/mole}$

$T_b=373.18K$

Therefore, the boiling point of a solution is, 373.18 K