Q38 / 160 · AP EAMCET 2011 · Chemistry
$19 \mathrm{~g}$ of a mixture containing $\mathrm{NaHCO}_3$ and $\mathrm{Na}_2 \mathrm{CO}_3$ on complete heating liberated $1.12 \mathrm{~L}$ of $\mathrm{CO}_2$ at STP. The weight of the remaining solid was $15.9 \mathrm{~g}$. What is the weight (in g) of $\mathrm{Na}_2 \mathrm{CO}_3$ in the mixture before heating?
- A8.4
- B15.9
- C4.0
- Correct answerD10.6
Solution
Molecular weight of
$\mathrm{NaHCO}_3=23+1+12+48=84$
Molecular weight of
$\mathrm{Na}_2 \mathrm{CO}_3=46+12+48=106$
Hence, total weight $=84+106=190$
$\because$ In $190 \mathrm{~g}$ of a mixture, weight of $\mathrm{Na}_2 \mathrm{CO}_3$ is
$=106$
$\therefore$ In $19 \mathrm{~g}$ of a mixture weight of
$\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 & =\frac{106 \times 19}{190} \\
& =10.6 \mathrm{~g}
\end{aligned}$
Hence, the correct answer is option: 'D'.