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The unit of ebullioscopic constant is
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Verified Answer
The correct answer is:
$\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$ or $\mathrm{K}$ (molality) ${ }^{-1}$
$\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$ or $\mathrm{K}$ (molality) ${ }^{-1}$
As we know from elevation in boiling point that
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$
$$
\mathrm{K}_{\mathrm{b}}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{m}}
$$
Unit of $K_b=\frac{\text { unit of } \Delta T_b}{\text { unit of } m}=\frac{K}{\text { molality }}$
$$
=\frac{\mathrm{K}}{\mathrm{mol} \mathrm{kg}^{-1}}=\mathrm{K} \mathrm{mol}^{-1} \mathrm{~kg}
$$
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$
$$
\mathrm{K}_{\mathrm{b}}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{m}}
$$
Unit of $K_b=\frac{\text { unit of } \Delta T_b}{\text { unit of } m}=\frac{K}{\text { molality }}$
$$
=\frac{\mathrm{K}}{\mathrm{mol} \mathrm{kg}^{-1}}=\mathrm{K} \mathrm{mol}^{-1} \mathrm{~kg}
$$
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