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  4. Aqueous solution of barium phosphate which is 100 % ionised has Δ Tf/Kf as 0.05. Hence, given solution is

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Q. Aqueous solution of barium phosphate which is $100\%$ ionised has $\Delta\,T_{f}/K_{f}$ as $0.05$. Hence, given solution is

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Solution:

$Ba_{3}(PO_{4})_{2} \rightleftharpoons 3\,Ba^{2+}+2\,PO^{3-}_{4}$
value of $i = 5 (100\%$ ionised)
so $\Delta\,T_{f}=iK_{f}\,m$
so $m=\frac{\Delta\,T_{f}}{ik_{f}}=\frac{0.05}{5}=0.01$