$6.2$ 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: $2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$ . Residue is dissolved in water and formed $100 m L$ solution and its $10 m L$ portion requires $7.5 m L$ of $0.2 M$ aqueous solution of $B a C l_{2}$ for complete precipitation of carbonates. Determine weight (in g) of $N a_{2} C O_{3}$ in the original sample.

1.59

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

1.06

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

0.53

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

None of these

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

Open in App

Solution

The correct option is B $1.06$
Loss in weight due to heating $= 6.2 \times \frac{5}{100} = 0.31$ g.
$2 N a H C O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$
$2 \times 84 g 44 + 18 g$ loss
$1$ mole $N a H C O_{3} \to \frac{62}{2} = 31 g$ loss due to heating
$0.31 g$ loss from $0.01$ mol 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 carbonates in original sample $= 0.015 - 0.005 = 0.001$
Mass of $N a_{2} C O_{3}$ in original sample $= 0.01 \times 106 = 1.06 g$

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).