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Aluminium phosphate is 100% ionised in 0.01 molal aqueous solution. Hence, ${ΔT}_{b} / K_{b}$ is:

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0.01

0.015

0.0175

0.02

answer is D.

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Detailed Solution

${AIPO}_{4} ⇌ {Al}^{3 +} + {PO}_{4}^{3 -}$

$\begin{array}{l} i = 2 \\ {ΔT}_{b} = i \times K_{b} \times m \\ ∴ \frac{{ΔT}_{b}}{K_{b}} = 0 .02 \end{array}$

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