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

Concentration Terms

12.8 g of mix...

Question

$12.8$ 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 :

ionic mass of

M^{+}

18

g/mol

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

mixture has equal mole fraction of

M C l

and

K C l

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

K C l

and

M C l

when separately treated with

A g N O_{3}

solution would give same number of moles of while precipitates

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

A g C l

is water soluble

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

Open in App

Solution

The correct options are
A ionic mass of $M^{+}$ is $18$ g/mol
B mixture has equal mole fraction of $M C l$ and $K C l$
C $K C l$ and $M C l$ when separately treated with $A g N O_{3}$ solution would give same number of moles of while precipitates
The reactions are as follows:
$M C l (M + 35.5) + A g N O_{3} ⟶ A g C l ↓ 143.5$
$K C l 74.5 + A g N O_{3} ⟶ A g C l ↓ 143.5$
Let $M C l = x g$ , then $K C l = (12.8 - x) g$

$A g C l d u e t o M C l = \frac{143.5 x}{(M + 35.5)}$

$A g C l d u e t o K C l = \frac{143.5}{74.5} (12.8 - x)$

$\frac{143.5 x}{(M + 35.5)} + \frac{143.5 (12.8 - x)}{74.5} = 28.7$

Since $M C l$ is volatile, in second case, $A g C l$ is due to $M C l$ only.

Thus, $[\frac{143.5 x}{M + 35.5}] = 14.35$

$∴ 14.35 + \frac{143.5 (12.8 - x)}{74.5} = 28.7$

$1 + \frac{10 (12.8 - x)}{74.5} = 2$

$(12.8 - x) = 7.45$

$x = 5.35$ g

Thus, $\frac{143.5 \times 5.35}{(M + 35.5)} = 14.35$
$∴ M = 18$

Thus, ionic mass of $M^{+} = 18$ g/mol. Hence, (a) is true and molar mass of $M C l = 53.5$ g/mol

Thus, moles of $M C l = \frac{5.35}{53.5} = 0.1$

Moles of $K C l = [\frac{12.80 - 5.35}{74.5}] = 0.1$

$A g C l$ is water insoluble.

Hence, mixture contains equal mole fraction of $M C l$ and $K C l$ .

Suggest Corrections

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 :

Q. 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:

When one mole of a complex is treated with excess of $A g N O_{3}$ solution, 143.5 g of white precipitate is formed, the complex is :

Q. A six-coordinate complex of formula

C r C l_{6} .6 H_{2}

O is green in colour. A

0.1 M

solution of the complex when treated with excess of

A g N O_{3}

gave

28.7 g

of white precipitate. The formula of the complex would be:

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) :

Join BYJU'S Learning Program

12.8 g of mixture of MCl (volatile) and KCl (non-volatile) on reaction with excess of aqueous AgNO3 solution gave 28.7 g white precipitate. 12.8 g of same mixture on a heating gave a gas that on passing into AgNO3 solution gave 14.35 g of white precipitate. Hence :

$12.8$ 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 :