A hydrated salt of ${N a}_{2} {S O}_{3}$ loses 22.22% of its mass on strong heating. The hydrated salt is :

{N a}_{2} {S O}_{3} . {4 H}_{2} O

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

{N a}_{2} {S O}_{3} . {6 H}_{2} O

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

{N a}_{2} {S O}_{3} . H_{2} O

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

{N a}_{2} {S O}_{3} . {2 H}_{2} O

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

Open in App

Solution

The correct option is D ${N a}_{2} {S O}_{3} . {2 H}_{2} O$
Let the hydrated salt be $N a_{2} S O_{3} . x H_{2} O$ .
On heating, it loses water.
So, it will lose $18 x$ g of water which is $22.22$ % of the total weight of the compound (given).
So, according to the given condition,
$\frac{18 x}{18 x + 126} \times 100 = 22.22$ ( $126$ is molecular mass of $N a_{2} S O_{3}$ )
$⟹ x = 2$
So, the hydrated salt is $N a_{2} S O_{3} .2 H_{2} O$ .

Suggest Corrections

Similar questions

{N a}_{2} {C O}_{3} . {x H}_{2} O

x

({N a}_{2} {S O}_{3} + K_{2} {C r}_{2} O_{7})

d i l . H_{2} {S O}_{4}

A l = 10.5 %, K = 15.1 %, S = 24.8 %

= 49.6 %

A l = 27, K = 39, S = 32

K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

K M n O_{4} + {N a}_{2} {S O}_{3} \to M n O_{2} + {N a}_{2} {S O}_{4} + K O H

C u + H {N O}_{3} \to C u {({N O}_{3})}_{2} + {N O}_{2} + H_{2} O

K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

Join BYJU'S Learning Program