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

Question

Aqueous solution of barium phosphate which is 100 % ionised has Δ Tf/Kf as 0.05. Hence, given solution is (A) 0.01 molal (B) 0.02 molal (C) 0.04 molal (D) 0.05 molal

Tardigrade Admin · Accepted Answer

Ba3(PO4)2 leftharpoons 3 Ba2++2 PO3-4 value of i = 5 (100 % ionised) so Δ Tf=iKf m so m=(Δ Tf/ikf)=(0.05/5)=0.01

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

0.01 molal

0.02 molal

0.04 molal

0.05 molal

$B a_{3} (P O_{4})_{2} ⇌ 3 B a^{2 +} + 2 P O_{4}^{3 -}$
value of $i = 5 (100%$ ionised)
so $Δ T_{f} = i K_{f} m$
so $m = \frac{Δ T _{f}}{i k _{f}} = \frac{0.05}{5} = 0.01$

Q. Aqueous solution of barium phosphate which is 100% ionised has ΔTf​/Kf​ as 0.05. Hence, given solution is