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Sea water contains 29.25% NaCl29.25 \% ~\mathrm{NaCl}29.25% NaCl and 19% MgCl219 \% ~\mathrm{MgCl}_{2}19% MgCl2 by weight of solution. The normal boiling point of the sea water is _____________ ∘C{ }^{\circ} \mathrm{C}∘C (Nearest integer)
Assume 100%100 \%100% ionization for both NaCl\mathrm{NaCl}NaCl and MgCl2\mathrm{MgCl}_{2}MgCl2
Given : Kb(H2O)=0.52 K kg mol−1\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}Kb(H2O)=0.52 K kg mol−1
Molar mass of NaCl\mathrm{NaCl}NaCl and MgCl2\mathrm{MgCl}_{2}MgCl2 is 58.5 and 95 g mol−1\mathrm{g} \mathrm{~mol}^{-1}g mol−1 respectively.
Amount of solvent =100−(29.25+19)=51.75 g100-(29.25+19)=51.75\mathrm{~g}\\\\100−(29.25+19)=51.75 g
Δ Tb=[2 × 29.25 × 100058.5 × 51.75+3 × 19 × 100095 × 51.75] × 0.52 \Delta \mathrm{T}_{\mathrm{b}}=\left[\frac{2 \times 29.25 \times 1000}{58.5 \times 51.75}+\frac{3 \times 19 \times 1000}{95 \times 51.75}\right] \times 0.52 \\Δ Tb=[58.5 × 51.752 × 29.25 × 1000+95 × 51.753 × 19 × 1000] × 0.52
Δ Tb=16.075 \Delta \mathrm{~Tb}=16.075 \\Δ Tb=16.075
Δ Tb=(Tb) solution −(Tb) solvent \Delta \mathrm{~Tb}=\left(\mathrm{T}_{\mathrm{b}}\right)_{\text {solution }}{ }{-}\left(\mathrm{T}_{\mathrm{b}}\right)_{\text {solvent }} \\Δ Tb=(Tb) solution −(Tb) solvent
(Tb) solution =100+16.07 \left(\mathrm{T}_{\mathrm{b}}\right)_{\text {solution }}=100+16.07 \\(Tb) solution =100+16.07
=116.07∘ C\quad=116.07^{\circ} \mathrm{C}=116.07∘ C
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