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Properties of Ionic compound

11.2 of the m...

Question

11.2 of the mixture of $M C l$ (volatile ) and $N a C l$ gave 28.7g of white ppt with an excess of $A g N O_{3}$ solution. 11.2 g of the same mixture on heating gave that on passing into $A g N O_{3}$ solution gave 14.35g of white ppt. Hence:

ionic mass of

M^{+}

is 18

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mixture has equal mol fraction of

M C l

and

N a C l

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M C l

and

N a C l

are in 1:2 molar ratio

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ionic mass of

M^{+}

is 10

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Solution

The correct options are
A ionic mass of $M^{+}$ is 18
B mixture has equal mol fraction of $M C l$ and $N a C l$
Comparing the $2$ experiments shows that exactly twice as much was produced by the mixture, as by $M C l$ alone. Since both formulas only contain $1$ atom of $C l$ , it shows that mixture contains an equal no.of moles of both compounds. So, their mole fraction would both be the answer to the problem, it is $(b)$ . So, $(c)$ is not possible.
However just to confirm $14.35 g A g C l = 0.01$ moles, so $11.2 g M C l$ must contain $0.01$ moles of $C l$ , or $3.545 g$ $
Hence, gram of $M = 11.2 - 3.545 = 7.655 = 0.1 m o l m^{+}$
So, $M^{+}$ weights $76.55 g$ . This rules out $(A)$ and $(B)$

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Similar questions

11.2

g of mixture of

M C l

(volatile) and

N a C l

gave

28.7

g of white precipitate with excess of

A g N O_{3}

solution.

11.2

g of same mixture on heating gave a gas that on passing into

A g N O_{3}

solution gave

14.35

g of white precipitate.

M C l

and

N a C l

have the molar ratio

x : y

.
The value of

10 (x + y)

is :

12.8

g of mixture of

M C l

(volatile) and

K C l

(non-volatile) on reaction with excess of aqueous

A g N O_{3}

solution gave

28.7

g white precipitate.

12.8

g of same mixture on a heating gave a gas that on passing into

A g N O_{3}

solution gave

14.35

g of white precipitate. Hence :

Q. A mixture contains

N a C l

and an unknown chloride

M C l

. (a)

1

g of this is dissolved in water, excess of acidified

A g N O_{3}

solution is added to it so that

2.567

g of white ppt. is obtained. (b)

1

g of original mixture is heated to

300

K. Some vapours come out which are absorbed in

A g N O_{3}

(acidified) solution.

1.341

g of white precipitate is obtained. The molar mass of the unknown chloride is (only integer part) :

Q. Colourless salt (A) give white ppt (B) with NaOH; ppt (B) dissolves in excess of NaOH to form (C). On passing

H_{2} S

gas into (C) white ppt (D) appears. (A) also gives white ppt (E) with

A g N O_{3}

solution. What is (A)?

10

mL of a solution of

K C l

containing

N a C l

on evaporation gave

0.93

g of the mixed salt, which gave

1.865

g of

A g C l

by the reaction with

A g N O_{3}

. The quantity of

N a C l

10

mL of the solution is (write it in multiples of

100

) :

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11.2 of the mixture of MCl (volatile ) and NaCl gave 28.7g of white ppt with an excess of AgNO3 solution. 11.2 g of the same mixture on heating gave that on passing into AgNO3 solution gave 14.35g of white ppt. Hence:

11.2 of the mixture of $M C l$ (volatile ) and $N a C l$ gave 28.7g of white ppt with an excess of $A g N O_{3}$ solution. 11.2 g of the same mixture on heating gave that on passing into $A g N O_{3}$ solution gave 14.35g of white ppt. Hence: