You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

Total molalit...

Question

Total molality after the addition of $H g (C N)_{2}$ into aqueous solution of $K C N$ .

(0.2842 - 0.095 m)

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

(0.4734 - 0.095 m)

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

0.2842

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

0.4734

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

Open in App

Solution

The correct option is B $(0.4734 - 0.095 m)$
The chemical reaction:
$H g (C N)_{2} + m C N^{-} \to H g (C N)_{m + 2}^{m^{-}}$
Initial moles $0.095$ $0.1892$ $0$
Final moles $0$ $0.1892 - 0.095$ $0.095$
$∴$ Total molality after addition of $H g (C N)_{2}$
$= molality of K^{+} + molality of C N^{-} + molality of H g (C N)_{m + 2}^{m^{-}}$
$= 0.1892 + (0.1892 - 0.0095 m) + 0.095$
$= 0.4734 - 0.095 m$

Suggest Corrections

Similar questions

The freezing point of an aqueous solution of

K C N

containing

0.1892

mol/kg H

_{2}

O was

- {0.704}^{o}

C. On adding

0.095

mole of

H g (C N)_{2}

, the freezing point of the solution was

- {0.53}^{o}

H g (C N)_{2}

is the limiting reactant.

H g (C N)_{2} + m C N^{-} \to H g (C N)_{m + 2}^{m -}

The value of m is?

Q. A zinc rod is dipped in

0.095 M

solution of

Z n S O_{4}

298 K

. Calculate the electrode potential of zinc electrode

(E_{Z a^{2 +} / Z a}^{0} = - 0.76 V)

Q. Aqueous solution of

F e S O_{4}

gives tests for both

F e^{2 +}

and

S O_{4}^{2 -}

but after addition of excess of KCN, solution ceases to give test for

F e^{2 +}

. This is due to the formation of :

Q. The freezing point of an aqueous solution of

K C N

containing

0.1892

mol kg

^{- 1}

was

- {0.704}^{o} C

. On adding

0.045

mole of

H g (C N)_{2}

, the freezing point of the solution was

- {0.620}^{o} C

. If whole of

H g (C N)_{2}

is used in complex formation according to the equation,

H g (C N)_{2} + m K C N \to K_{m} [H g (C N)_{m + 2}]

Assume

[H g (C N)_{m + 2}]^{m -}

is not ionised and the complex molecule is 100% ionised.

K_{f} (H_{2} O)

1.86

kg mol

^{- 1}

K.
The formula of the complex is :

Q. When a metal rod

M

is dipped into an aqueous colourless concentrated solution of

A g N O_{3}

, the solution turns light blue. Addition of aqueous

N a C l

to the blue solution gives a white precipitate O. Addition of aqueous

N H_{3}

dissolves

O

and gives an intense blue solution.
The final solution contains:

Join BYJU'S Learning Program

Total molality after the addition of Hg(CN)2 into aqueous solution of KCN.

Total molality after the addition of $H g (C N)_{2}$ into aqueous solution of $K C N$ .