1 × 10-3 m solution of Pt ( NH 3)2 Cl 4 in H 2 O shows depression in freezing point by 0.0054° C. The ionisable Cl -ions will be (Given, Kf( H 2 O )=1.860 K kg mol -1 )

Question

1 × 10-3 m solution of Pt ( NH 3)2 Cl 4 in H 2 O shows depression in freezing point by 0.0054° C. The ionisable Cl -ions will be (Given, Kf( H 2 O )=1.860 K kg mol -1 ) (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Δ Tf=i Kf m ⇒ 0.0054=i × 1.86 × 0.001 i=(5.4/1.86) ≈ 3 [ Pt ( NH 3)2 Cl 2] Cl 2 leftharpoons[ Pt ( NH 3)2 Cl 2]2++2 Cl - So, ionisable Cl -ions are 2 .

Q. $1 \times 1 0^{- 3} m$ solution of $Pt (N H_{3})_{2} C l_{4}$ in $H_{2} O$ shows depression in freezing point by $0.005 4^{\circ} C$ . The ionisable $C l^{-}$ ions will be (Given, $K_{f} (H_{2} O) = 1.860 K k g m o l^{- 1}$ )

$1$

$2$

$3$

$4$

$Δ T_{f} = i K_{f} m$
$\Rightarrow 0.0054 = i \times 1.86 \times 0.001$
$i = \frac{5.4}{1.86} \approx 3$
$[Pt (N H_{3})_{2} C l_{2}] C l_{2} ⇌ [Pt (N H_{3})_{2} C l_{2}]^{2 +} + 2 C l^{-}$
So, ionisable $C l^{-}$ ions are $2$ .

Q. 1×10−3m solution of Pt(NH3​)2​Cl4​ in H2​O shows depression in freezing point by 0.0054∘C. The ionisable Cl−ions will be (Given, Kf​(H2​O)=1.860Kkgmol−1 )

1

2

3

4

ΔTf​=iKf​m ⇒0.0054=i×1.86×0.001 i=1.865.4​≈3 [Pt(NH3​)2​Cl2​]Cl2​⇌[Pt(NH3​)2​Cl2​]2++2Cl− So, ionisable Cl−ions are 2 .

Q. $1 \times 1 0^{- 3} m$ solution of $Pt (N H_{3})_{2} C l_{4}$ in $H_{2} O$ shows depression in freezing point by $0.005 4^{\circ} C$ . The ionisable $C l^{-}$ ions will be (Given, $K_{f} (H_{2} O) = 1.860 K k g m o l^{- 1}$ )

$1$

$2$

$3$

$4$

$Δ T_{f} = i K_{f} m$
$\Rightarrow 0.0054 = i \times 1.86 \times 0.001$
$i = \frac{5.4}{1.86} \approx 3$
$[Pt (N H_{3})_{2} C l_{2}] C l_{2} ⇌ [Pt (N H_{3})_{2} C l_{2}]^{2 +} + 2 C l^{-}$
So, ionisable $C l^{-}$ ions are $2$ .