$6.2$ g of a sample containing $N a_{2} C O_{3}, N a H C O_{3}$ amd non-volatile inert impurity on heating loses $5 %$ of its weight due to formation of $H_{2} O$ and $C O_{2}$ as in reaction:
$2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$
Residue is dissolved in water and formed $100$ ml solution and its $10$ ml portion requires $4.5$ ml of $0.2 M$ aqueous solution of $B a C l_{2}$ for complete precipitation of carbonates. If number of moles of $N a_{2} C O_{3}$ in the original sample were $q$ , then the value of $100 q$ is:

2

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

1

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

6

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

5

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

Open in App

Solution

The correct option is B $1$
The reaction is as follows:
$2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$
Loss in weight due to heating $= 6.2 \times \frac{5}{100} = 0.31$ g.
$0.31$ g loss from $0.01$ mol (or $6.2$ g) of $N a H C O_{3}$ .
Moles of $N a_{2} C O_{3}$ produced $= \frac{0.01}{2} = 0.005$
Total moles of carbonate reacted with $B a C l_{2}$ $= (7.5 \times 0.2 \times \frac{100}{10}) \times 10^{- 3} = 0.015$ .
Moles of carbonate in original sample, $q$ $=$ $0.015 - 0.005$ $=$ $0.01$ .
Hence, $100 q = 100 \times 0.01$ $= 1$ .

Suggest Corrections

Similar questions

$6.2$ g of a sample containing $N a_{2} C O_{3}, N a H C O_{3}$ and non-volatile inert impurity on gentle heating loses $5 %$ of its weight due to reaction as shown below:

$2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$

Residue is dissolved in water and formed $100$ mL solution and its $10$ mL portion requires $7.5$ mL of $0.2 M$ aqueous solution of $B a C l_{2}$ for complete precipitation of carbonates. Determine the weight (in g) of $N a_{2} C O_{3}$ in the original sample (as nearest integer).