Calculate the boiling point of a one molar aqueous solution (density $=1.04 \mathrm{~gmL}^{-1}$ ) of potassium chloride, $\mathrm{K}_{\mathrm{b}}$ for water $=0.52 \mathrm{~kg}$ $\mathrm{mol}^{-1} .$ Atomic mass of $\mathrm{K}=39, \mathrm{Cl}=35.5$

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Aicharya · Accepted Answer

Volume of solution $=1000 \mathrm{~ml}$
Mass of the solution $=\mathrm{V} \times \mathrm{d}$
$=1000 \mathrm{~mL} \times 1.04 \mathrm{~g} / \mathrm{mL}=1040 \mathrm{~g}$
Amount of solute in $1000 \mathrm{~mL}$ solution
$=1 \mathrm{~M}$
$=1 \mathrm{~M} \times$ Molecular Mass $=1 \mathrm{M} \times 74.5 \mathrm{~g} / \mathrm{mol}$
$=74.5 \mathrm{~g}$
Mass of water $=$ Mass of solution $-$ Mass of
$\mathrm{KCl}=1040 \mathrm{~g}-74.5 \mathrm{~g}=965.5 \mathrm{~g}$
Molality of the solution $=\mathrm{m}=$
$\frac{\text { Moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\frac{1 \mathrm{~mol}}{(965.5 / 1000) \mathrm{kg}}$
$=1.0357$
$\Delta \mathrm{T}_{\mathrm{b}} \mathrm{i} . \mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}=2 \times 0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.0357$
$=1.078^{\circ} \mathrm{C}$.
Boiling point of the solution
$=100^{\circ} \mathrm{C}+1.078^{\circ} \mathrm{C}=101.078^{\circ} \mathrm{C}$

Calculate the boiling point of a one molar aqueous solution (density $=1.04 \mathrm{~gmL}^{-1}$ ) of…

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