Question

Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml solution

• A. 2.65 grams
• B. 4.95 grams
• C. 6.25 grams
• D. None of these

Answer 1

Answer: (A)

We know that
Molarity $=\frac{W}{M\times V}$
where;
W= Mass of $N{a}_{2}C{O}_3$ in gram
M = Molecular mass of $N{a}_{2}C{O}_3$ in grams $=106$
V =Volume of solution in litres
$=\frac{250}{1000}=0.25$
Molarity $=\frac{1}{10}$
Hence, $=\frac{1}{10}=\frac{W}{106\times 0.25}$
\ or $W=\frac{106\times 0.25}{10}$
$=2.65$ grams