Question

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol −1 .

Answer 1

mass of water (solvent ) , W = 500g
Boiling point of water , Tb = 99.63°C
molal elevation constant for water, Kb= 0.52K.kg/mol
elevation in boiling point , ∆Tb = 100°C - 99.63°C = 0.37°C/0.37K
molar mass of sucrose $C_{12}H_{22}O_{11}$,M=342g/mol
Let mass of sucrose is w g

we know, $\Delta{T_b}=\frac{K_b\times1000\times w}{M\times W}$

0.37 = (0.52 × 1000 × w)/(342 × 500)
0.37 = (0.52 × 2 × w)/(342)
w = 342 × 0.37/(1.04) = 121.67g

hence, weight of sucrose is 121.67g

Answer 2

Answer :

Given :-

Mass of water, $W_{A}$ = 500 g

Boiling point of water = 100°C

Molal elevation constant, $K_{b}$ = 0.52 K kg $mol^{-1}$

Elevation of boiling point, ∆$T_{b}$ = $T_{b} - T_{b}°$ = 100 − 99.63 = 0.37

Molecular weight of $C_{12}H_{22}O_{11}$ (sucrose),$M_{B}$ = 11 × 12 + 22 × 1 + 11 × 16 = 342 g $mol^{-1}$

To find,

Weight of Sucrose = $W_{B}$

Using the formula,

∴ Δ$T_{b}$ = $\frac{W_{B}}{M_{B}} × \frac{1000}{W_{A}} × K_{b}$

⇒ 0.37 = $\frac{W_{B}}{342} × \frac{1000}{500} × 0.52$

⇒ $W_{B} = \frac{0.37 × 171}{0.52}$

⇒ $W_{B}$ = 121.67 g