Question

When 0.6 g of urea dissolved in 100 g of water the water will boil

Answer 1

Answer:

The boiling point of the solution is 100.052°C.

Explanation:

Mass of the uera -= 0.6 g

Mass of the water = 100g =0.1 kg

Molality of the solution = $\frac{0.6 g}{60 g/mol\times 0.1 kg}=0.1 m$

Elevation in boiling point = $\Delta T_b$

Boiling point of solution = $T_b$

$T_b=K_b\times m=0.52 K kg/mol\times 0.1 =0.052 K$

Boiling point of pure water = T =373 K

$\Delta T_b=T_b-T$

$T_b=373 K+0.052 K=373.052 K=100.052^oC$

The boiling point of the solution is 100.052°C.