A 0-001 molal solution of -PtNH34cl4- in water had a freezing point depression of 0-0054- If Kf for water is 1-80- the correct formulation for the above molecule is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

A 0.001 molal...

Question

A 0.001 molal solution of $[P t (N H_{3})_{4} c l_{4}]$ in water had a freezing point depression of ${0.0054}^{\circ}$ . If $K_{f}$ for water is 1.80, the correct formulation for the above molecule is

[P t (N H_{3})_{4} C l_{3}] C l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[P t (N H_{3})_{4}] C l_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[P t (N H_{3})_{4} C l_{2}] C l_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[P t (N H_{3})_{4} C l_{2}] C l_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $[P t (N H_{3})_{4} C l_{2}] C l_{2}$
$△ T = i K_{f} m$
$\Rightarrow i = \frac{△ T_{f}}{K_{f} m} = \frac{0.0054}{1.80 \times 0.001} = 3$
Thus complex produces three ions per molecule. hence possible correct formula is
$[P t (N H_{3})_{4} C l_{2}] C l_{2}$

Suggest Corrections

Similar questions

Q. A 0.001 molal solution of

[P t (N H_{3})_{4} c l_{4}]

in water had a freezing point depression of

{0.0054}^{\circ}

. If

K_{f}

for water is 1.80, the correct formulation for the above molecule is

A 0.001 molal solution of in water had a freezing point depression of 0.0054°C. If K_f for water is 1.80, the correct formulation for the above molecule is

Q. A 0.001 molal solution of

[P t (N H_{3})_{4} C l_{4}]

in water has a freezing point depression of

{0.0054}^{0} C

. If

K_{f}

for water is 1.80 K.kg.mol

^{- 1}

, the correct formulation of the above molecule is:

0.001

molal solution of

[P t {({N H}_{3})}_{4} {C l}_{4}]

in water had a freezing point depression of

0.0054^{o} C

. If

K_{f}

for water is

1.80

, the correct formula of the compound is :

Q. A 0.001 molal solution of a complex represented as

[P t (N H_{3})_{4} C l_{4}]

in water had freezing point depression of

{0.0053}^{o} C

. Given,

K_{f}

for

H_{2} O = 1.86 K m o l a l i t y^{- 1}

. Assuming 100% ionisation of the complex, write the ionisation nature and formula of complex.

Join BYJU'S Learning Program

A 0.001 molal solution of [Pt(NH3)4cl4] in water had a freezing point depression of 0.0054∘. If Kf for water is 1.80, the correct formulation for the above molecule is

The correct option is D [Pt(NH3)4Cl2]Cl2△T=i Kf m ⇒ i=△TfKf m=0.00541.80×0.001=3 Thus complex produces three ions per molecule. hence possible correct formula is [Pt(NH3)4Cl2]Cl2

A 0.001 molal solution of $[P t (N H_{3})_{4} c l_{4}]$ in water had a freezing point depression of ${0.0054}^{\circ}$ . If $K_{f}$ for water is 1.80, the correct formulation for the above molecule is

The correct option is D $[P t (N H_{3})_{4} C l_{2}] C l_{2}$
$△ T = i K_{f} m$
$\Rightarrow i = \frac{△ T_{f}}{K_{f} m} = \frac{0.0054}{1.80 \times 0.001} = 3$
Thus complex produces three ions per molecule. hence possible correct formula is
$[P t (N H_{3})_{4} C l_{2}] C l_{2}$