Question

Calculate the boiling point of solution when 4 g of mgso4 (m=120 g mol-1 ) was dissolved in 100 g of water, assuming mgso4 undergoes complete ionization. (kb for water = 0.52 k kg mol-1 )

Answer 1

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Answer 2

Answer: The boiling point of the solution is $373.35K$.

Explanation:

Weight of solvent (water) = 100 g = 0.1 kg (1 kg=1000 g)

Molar mass of $MgSO_4$ =  120 g/mol

Mass of $MgSO_4$ added = 4 g

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

i = Van'T Hoff factor

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

$\alpha =\frac{i-1}{n-1}=100\%=1.0=\frac{i-1}{2-1}$

i = 2

$\Delta T_b=2\times 0.52 K kg/mol\frac{4g g}{120g/mol\times 0.1kg}$

$\Delta T_b=T_b-T^{o}_b=T_b-373K$

$(T_b-373)K =2\times 0.52 K kg/mol\frac{4g g}{120g/mol\times 0.1kg}$

$T_b=373.35K$

The boiling point of the solution is $373.35K$.