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Byju's Answer

Standard XII

Chemistry

Central Atom and Ligands

3.24 g of H...

Question

3.24 g of $H g (N O_{3})_{2} (m o l a r m a s s = 324)$ dissolved in 1000 g of water constitutes a solution having a freezing point of $- {0.0558}^{o} C$ while 21.68 g of $H g C l_{2}$ (Molar mass $=$ 271) in 2000 g of water constitutes a solution with a freezing point of $- {0.0744}^{o} C$ . The $K_{f}$ for water is $1.86 \frac{K . K g}{m o l}$ . About the state of ionization of these two solids in water it can be inferred that

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

and

H g C l_{2}

both are completely ionized

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

is fully ionized but

H g C l_{2}

is fully unionized

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

and

H g C l_{2}

both are completely unionized

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

is fully unionized but

H g C l_{2}

is fully ionized

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Solution

The correct option is B $H g (N O_{3})_{2}$ is fully ionized but $H g C l_{2}$ is fully unionized
We know,

$Δ T_{f} = i \times m \times K_{f}$

Now,

$0.0558 = i \times \frac{3.24}{324} \times \frac{1000}{1000} \times 1.86 \Rightarrow i = 3$ (100% dissociated)
$0.0744 = i \times \frac{21.68}{271} \times \frac{1000}{2000} \times 1.86 \Rightarrow i ≃ 1$ (almost undissociated)

Suggest Corrections

Similar questions

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. 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. 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]

Q. A certain mass of substance, when dissolved in 100 g

C_{6} H_{6},

lowers the freezing point by

{1.28}^{\circ} C .

The same mass of solute dissolved in 100 g water lowers the freezing point by

{1.40}^{\circ} C .

If the substance has normal molecular weight in benzene and is completely ionized in water, into how many ions does it dissociate in water?

K_{f}

for

H_{2} O

and

C_{6} H_{6} O

are 1.86 and 5.12 K kg

m o l^{- 1} .

0.01 N

solution of

K C l

and

B a C l_{2}

are prepared in water. The freezing point of

K C l

is found to be

- 2^{\circ} C

. What is the freezing point of

B a C l_{2}

solution assuming both

K C l

and

B a C l_{2}

to be completely ionized?

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The correct option is B Hg(NO3)2 is fully ionized but HgCl2 is fully unionizedWe know,ΔTf=i×m×KfNow,0.0558=i×3.24324×10001000×1.86⇒i=3 (100% dissociated)0.0744=i×21.68271×10002000×1.86⇒i≃1 (almost undissociated)