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$19 \mathrm{~g}$ 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?
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Verified Answer
The correct answer is:
10.6
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
$\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 & =\frac{106 \times 19}{190} \\
& =10.6 \mathrm{~g}
\end{aligned}$
$\mathrm{NaHCO}_3=23+1+12+48=84$
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
$\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 & =\frac{106 \times 19}{190} \\
& =10.6 \mathrm{~g}
\end{aligned}$
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