Thank you for reporting, we will resolve it shortly
Q.
$250\, mL$ of sodium carbonate solution contains $2.65\, g$ of $Na_2CO_3$. If $10\, mL$ of this solution is diluted to $500\, mL$, the concentration of the diluted acid will be
Solutions
Solution:
Molarity of $Na_2CO_3$ solution
$= \frac{2.65}{106} \times \frac{1000}{250} = 0.1\,M$
$10 \,mL$ of this solution is diluted to $500\, mL$
$M_{1}V_{1} = M_{2}V_{2}$
$0.1 \times 10 = M_{2} \times 500$
$\Rightarrow M_{2} = 0.002\,M$