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

Byju's Answer

Standard XII

Chemistry

Van't Hoff Factor

For a solutio...

Question

For a solution of 0.849 g of mercurous chloride in 50g of $H g C l_{2} (l)$ , the freezing point depression is ${1.24}^{o} C . K_{f}$ for $H g C l_{2}$ is 34.3. What is the state of mercurous chloride in $H g C l_{2}$ ?
$[H g - 200 g, C l - 35.5 g]$

H g_{2} C l_{2}

molecules

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

H g C l

molecules

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

H g^{+}

and

C l^{-}

ions

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

H g_{2}^{2 +}

and

C l^{-}

ions

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

Open in App

Solution

The correct option is B As $H g_{2} C l_{2}$ molecules
The expression for the depression in the freezing point and the molality is $K_{f} = \frac{Δ T_{f}}{m}$

The molality of the solution is $\frac{\frac{0.849}{M}}{0.050}$
$1.24 = 34.3 [\frac{\frac{0.849}{M}}{0.050}]$

Hence, $M = 469.68$

Thus mercurous chloride is present in the form of $H g_{2} C l_{2}$ .

Suggest Corrections

Similar questions

Q. The bond of

H g - C l

is more ionic in

H g C l_{2}

than that in

H g_{2} C l_{2}

Q. The equilibrium constant for the disproportionation of

H g C l_{2}

into

H g C l^{+}

and

H g C l_{3}^{-}

is ...............
Given,

H g C l^{+} + C l^{-} ⇌ H g C l_{2}; K_{1} = 3 \times 10^{6}

H g C l_{2} + C l^{-} ⇌ H g C l_{3}^{-}; K_{2} = 9.0

Q. When

3.24 g

of mercuric nitrate

H g (N O_{3})_{2}

dissolved in

1 k g

of water, the freezing point of the solution is found to be

- {0.0558}^{\circ} C

. When

10.84 g

of mercuric chloride

H g C l_{2}

is dissolved in

1 k g

of water, the freezing point of the solution is

- {0.0744}^{\circ} C

K_{f} = 1.86 m o l^{- 1} K k g

. Will either of these dissociate into ions in an aqueous solution?

Q. When 3.24 g of mercuric nitrate,

H g (N O_{3})_{2}

is dissolved in 1000 g of water, the freezing point of the solution is found to be

- {0.0558}^{0} C

. But when 10.84 g of

H g C l_{2}

is dissolved in 1000 g of water, the freezing point depression is

{0.0744}^{0} C

K_{f}

for water is

{1.86}^{0} C m o l a l^{- 1} .

Now, select the correct statement(s). (Given:` Atomic mass of Hg = 200)

Q. Given the equilibrium constants

H g C l^{\oplus} + C l^{⊖} \to H g C l_{2}, K_{1} = 3 \times 10^{6}

H g C l_{2} + C l^{⊖} \to H g C l_{3}^{⊖}, K_{2} = 8.9

The equilibrium constant for the disproportionation equilibrium

2 H g C l_{2} \to H g C l^{\oplus} + H g C l_{3}^{⊖}

is :

Join BYJU'S Learning Program

For a solution of 0.849 g of mercurous chloride in 50g of HgCl2(l), the freezing point depression is 1.24oC.Kf for HgCl2 is 34.3. What is the state of mercurous chloride in HgCl2? [Hg−200 g,Cl−35.5 g]

The correct option is B As Hg2Cl2 moleculesThe expression for the depression in the freezing point and the molality is Kf=ΔTfmThe molality of the solution is 0.849M0.0501.24=34.3[0.849M0.050]Hence, M=469.68Thus mercurous chloride is present in the form of Hg2Cl2.

For a solution of 0.849 g of mercurous chloride in 50g of $H g C l_{2} (l)$ , the freezing point depression is ${1.24}^{o} C . K_{f}$ for $H g C l_{2}$ is 34.3. What is the state of mercurous chloride in $H g C l_{2}$ ?
$[H g - 200 g, C l - 35.5 g]$