Question

$19\, g$ of a mixture containing $NaHCO _{3}$ and $Na _{2} CO _{3}$ on complete heating liberated $1.12 \,L$ of $CO _{2}$ at $STP$. The weight of the remaining solid was $15.9\, g$. What is the weight (in $g$ ) of $Na _{2} CO _{3}$ in the mixture before heating?

• A. 8.4
• B. 15.9
• C. 4.0
• D. 10.6

Answer 1

Answer: (D)

Molecular weight of

$NaHCO _{3}=23+1+12+48=84$

Molecular weight of

$Na _{2} CO _{3}=46+12+48=106$

Hence, total weight $=84+106=190$

$\because$ In $19\, g$ of a mixture, weight of $Na _{2} CO _{3}$ is $=106$

$\therefore $ In $19\, g$ of a mixture weight of

$Na _{2} CO _{3} =\frac{106 \times 19}{190} $

$=10.6 \,g$