Search any question & find its solution
Question:
Answered & Verified by Expert
The volume of $\mathrm{CO}_{2}\left(\mathrm{in} \mathrm{cm}^{3}\right)$ liberated at STP, when $1.06 \mathrm{~g}$ of anhydrous sodium carbonate is treated with excess of dilute $\mathrm{HCl}$ is [atomic mass of $\mathrm{Na}=23]$
Options:
Solution:
1455 Upvotes
Verified Answer
The correct answer is:
224
$\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl}$ (dil.) $\longrightarrow 2 \mathrm{NaCl}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
$1 \mathrm{~mol} \quad$ (dil.) $1 \mathrm{~mol}$
$=1.06 \mathrm{~g} \quad=22400 \mathrm{~cm}^{3}$
$1.06 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}$ gives $=22400 \mathrm{~cm}^{3}$ of CO$_{2}$
$\therefore 1.06$ g gives $=\frac{1.06 \times 22400}{106}=224 \mathrm{~cm}^{3}$ of $\mathrm{CO}_{2}$
$1 \mathrm{~mol} \quad$ (dil.) $1 \mathrm{~mol}$
$=1.06 \mathrm{~g} \quad=22400 \mathrm{~cm}^{3}$
$1.06 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}$ gives $=22400 \mathrm{~cm}^{3}$ of CO$_{2}$
$\therefore 1.06$ g gives $=\frac{1.06 \times 22400}{106}=224 \mathrm{~cm}^{3}$ of $\mathrm{CO}_{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.