JEE MAIN - Chemistry (2023 - 13th April Evening Shift - No. 18)
Sea water contains $$29.25 \% ~\mathrm{NaCl}$$ and $$19 \% ~\mathrm{MgCl}_{2}$$ by weight of solution. The normal boiling point of the sea water is _____________ $${ }^{\circ} \mathrm{C}$$ (Nearest integer)
Assume $$100 \%$$ ionization for both $$\mathrm{NaCl}$$ and $$\mathrm{MgCl}_{2}$$
Given : $$\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$
Molar mass of $$\mathrm{NaCl}$$ and $$\mathrm{MgCl}_{2}$$ is 58.5 and 95 $$\mathrm{g} \mathrm{~mol}^{-1}$$ respectively.
Answer
116
Explanation
Amount of solvent = 100 - (29.25 + 19) = 51.75 g
Now, we can calculate the boiling point elevation using the given formula:
ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52
ΔTb = 16.075
The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:
Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C
Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.
Now, we can calculate the boiling point elevation using the given formula:
ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52
ΔTb = 16.075
The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:
Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C
Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.
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