Question

$250ml$ of a sodium carbonate contains $2.65$ grams ${Na}_2{CO}_3$. If $10ml$ of that solution is diluted to one litre, what is the concentration of the resultant solution (mol wt. of ${Na}_2{CO}_3=106$)

• A. $0.1M$
• B. $0.001M$
• C. $0.01M$
• D. ${10}^{-4}M$

Answer 1

Answer: (B)

$0.001M$

Given, $250ml$ $N{a}_{2}C{O}_3$ solution contains $2.65g$ $N{a}_{2}C{O}_3$

$\Rightarrow$ Number of moles of $N{a}_{2}C{O}_3$ in the solution.

$=\frac{2.65}{106}=0.025$ moles $[\because Mole=\frac{Wt.}{G.Wt}]$

Now, Molarity of given solution,

$M=\frac{No.ofmoles}{VinL}=\frac{0.025}{250\times {10}^{-3}}=0.1M$

Now, to calculate no. of moles of $N{a}_{2}C{O}_{3}in$10ml$ solution

$\Rightarrow \frac{No.ofmoles}{VinL}$

$\Rightarrow$ No. of moles=$M\times V$ in $L$

$=0.1\times 10\times {10}^{-3}$

$=0.001$ moles

Finally, for molarity calculation of final solution i.e, $1L$ solution

Molarity=$\frac{No.ofmoles}{VinL}$

$=\frac{0.001}{1}$

$=0.001M$.