Question

Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml of 0.1 M solution:-

• A. 2.65 gm
• B. 4.95 gm
  
• C. 6.25 gm
  
• D. None

Answer 1

The correct option is A 2.65 gm

Molarity =$\frac{W}{M\times v}$

W=Mass of $N{a}_{2}C{O}_3$ in gram

M=Molecular mass of $N{a}_{2}C{O}_3$ in gram = 106

V=Volume of solution in litres = $\frac{250}{1000}=0.25$

Molarity =$\frac{1}{10}$

Hence, =$\frac{1}{10}=\frac{W}{106\times 0.25}\Rightarrow W=\frac{106\times 0.25}{10}=2.65$ gram