Question 14 Marks
Calculate the boiling point of solution when 2 g of $ Na_{2}SO_{4} $ (m = 142 g $ Mol^{-1} $) was dissolved in 50 g of water, assuming $ Na_{2}SO_{4} $ undergoes complete ionization?
Answer
View full question & answer→$M =142 g mol ^{-1}$
$M _{ B }=142 \times 10^{-3} kg mol ^{-1}$
$W _{ A }=50 g=50 \times 10^{-3} kg$
As we know that
$\Delta T _b=\frac{ K _b \times W _{ B } \times 1000}{ M _{ B } W _{ A }} \times i$
$=\frac{0.52 \times 2 \times 10^{-3} \times 1000}{142 \times 10^{-3} \times 50 \times 10^{-3}} \times 3$
[For complete ionization $i=n, NaSO _4 \Rightarrow i=3$ ]
$=\frac{1.04 \times 1000 \times 1000 \times 3}{142 \times 50}$
$=\frac{3 \times 20.8 \times 100}{142}$
$=3 \times 0.146 \times 1000$
$\Delta T _b=3 \times 146 K=438 K$
Boiling point of solution $( T )= T _0+\Delta T _b$
$=373+438 K=811 K$
$M _{ B }=142 \times 10^{-3} kg mol ^{-1}$
$W _{ A }=50 g=50 \times 10^{-3} kg$
As we know that
$\Delta T _b=\frac{ K _b \times W _{ B } \times 1000}{ M _{ B } W _{ A }} \times i$
$=\frac{0.52 \times 2 \times 10^{-3} \times 1000}{142 \times 10^{-3} \times 50 \times 10^{-3}} \times 3$
[For complete ionization $i=n, NaSO _4 \Rightarrow i=3$ ]
$=\frac{1.04 \times 1000 \times 1000 \times 3}{142 \times 50}$
$=\frac{3 \times 20.8 \times 100}{142}$
$=3 \times 0.146 \times 1000$
$\Delta T _b=3 \times 146 K=438 K$
Boiling point of solution $( T )= T _0+\Delta T _b$
$=373+438 K=811 K$