##### Exercise | Q 11 | Page 46

# A solution of citric acid C6H8O7 in 50 g of acetic acid has a boiling point elevation of 1.76 K. If Kb for acetic acid is 3.07 K kg mol-1, what is the molality of solution?

### Solution:

**Given:** Boiling point elevation =ΔT_{b} = 1.76 K

K_{b} of acetic acid = 3.07 K kg mol^{-1}

Mass of acetic acid = 50 g

**To find:** Molality of the solution

**Formula:** ΔT_{b} = K_{b} m

**Calculation:** Using formula and rearranging, we get,

$\text{m}=\frac{\u25b3{\text{T}}_{\text{b}}}{{\text{K}}_{\text{b}}}=\frac{1.76\text{K}}{3.07{\text{K kg mol}}^{-1}}=0.573{\text{mol kg}}^{-1}$ = 0.573 m

The molality of the solution is 0.573 m.

**Given:** Boiling point elevation =ΔT_{b} = 1.76 K

K_{b} of acetic acid = 3.07 K kg mol^{-1}

Mass of acetic acid = 50 g

**To find:** Molality of the solution

**Formula:** ΔT_{b} = K_{b} m

**Calculation:** Using formula and rearranging, we get,

The molality of the solution is 0.573 m.