Question

Calculate the boiling point of solution when 4g of mgso4 was dissolved in 100g of water

Answer 1

Answer: The boiling point of solution is $100.31^0C$

Explanation:

Elevation in boiling point:

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

where,

$\Delta T_b$ = change in boiling point

i= vant hoff factor

$k_b$ = boiling point constant

m = molality

$T_{solution}-T_{solvent}=i\times k_b\times \frac{\text{ Moles of solute}\times 1000}{\text{ Mass of solvent in g}}$

moles of solute=$\frac{\text {given mass}}{\text {Molar mass}}=\frac{4g}{120g/mol}=0.03moles$

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

As it produces two ions on dissociation, i= 2

$T_{solution}-100^0C=2\times 0.512\times \frac{0.03\times 1000}{100g}$

$T_{solution}=100.31^0C$

The boiling point of solution is $100.31^0C$