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

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

Solutions

Solution:

We know that
Molarity $=\frac{W}{M \times V}$
where;
W= Mass of $Na_{2}CO_{3}$ in gram
M = Molecular mass of $Na_{2}CO_{3}$ 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