Question

boiling point of 10% aqueous solution of common salt at atmospheric pressure is​

Answer 1

Answer:  $102^0C$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

i = vant hoff factor = 2 (for NaCl as it dissociates to give 2 ions)

$\Delta T_b=T_b-T_b^0=(T_b-100)^0C$ = elevation in boiling point

$K_b$ = boiling point constant = $0.52K/kgmol$

m= molality

$\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Given :

10 g of $NaCl$ is dissolved in 100 g of solution

Weight of solvent (water)= (100-10)g = 90 g = 0.09 kg

Molar mass of solute (NaCl) = 58.5 g/mol

Mass of solute (NaCl) = 10 g

$(T_b-100)^0C=2\times 0.52\times \frac{10g}{58.5g/mol\times 0.09kg}$

$T_b=102^0C$

Thus value of boiling point of solution at 1 atm pressure is $102^0C$