PtCl4 . 6H2O can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of 3.72° Assume 100% ionisation and Kf(H2O)= 1.86° mol-1 kg, then complex is

Question

PtCl4 . 6H2O can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of 3.72° Assume 100% ionisation and Kf(H2O)= 1.86° mol-1 kg, then complex is (A) [Pt(H2O)6]Cl4 (B) [Pt(H2O)4Cl2]Cl2 . 2H2O (C) [Pt(H2O)3Cl3] Cl.3H2O (D) [Pt(H2O)2Cl4].4H2O

Tardigrade Admin · Accepted Answer

Δ Tf=i × m × Kf i=(3.72/1.86 × 1)=2 So salt is (Pt (H2O)3 Cl3) Cl. 3H2O

Q. $PtC l_{4} .6 H_{2} O$ can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of $3.7 2^{\circ}$ Assume 100% ionisation and $K_{f} (H_{2} O) = 1.8 6^{\circ} m o l^{- 1} k g$ , then complex is

$[Pt (H_{2} O)_{6}] C l_{4}$

$[Pt (H_{2} O)_{4} C l_{2}] C l_{2} .2 H_{2} O$

$[Pt (H_{2} O)_{3} C l_{3}] Cl .3 H_{2} O$

$[Pt (H_{2} O)_{2} C l_{4}] .4 H_{2} O$

$Δ T_{f} = i \times m \times K_{f}$
$i = \frac{3.72}{1.86 \times 1} = 2$
So salt is (Pt $(H_{2} O)_{3} C l_{3}) Cl .3 H_{2} O$

Q. PtCl4​.6H2​O can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of 3.72∘ Assume 100% ionisation and Kf​(H2​O)=1.86∘mol−1kg, then complex is

[Pt(H2​O)6​]Cl4​

[Pt(H2​O)4​Cl2​]Cl2​.2H2​O

[Pt(H2​O)3​Cl3​]Cl.3H2​O

[Pt(H2​O)2​Cl4​].4H2​O