250 mL of sodium carbonate solution contains 2.65 g of Na2CO3. If 10 mL of this solution is diluted to 500 mL, the concentration of the diluted acid will be

Question

250 mL of sodium carbonate solution contains 2.65 g of Na2CO3. If 10 mL of this solution is diluted to 500 mL, the concentration of the diluted acid will be (A) 0.01 M (B) 0.001 M (C) 0.05 M (D) 0.002 M

Tardigrade Admin · Accepted Answer

Molarity of Na2CO3 solution = (2.65/106) × (1000/250) = 0.1 M 10 mL of this solution is diluted to 500 mL M1V1 = M2V2 0.1 × 10 = M2 × 500 ⇒ M2 = 0.002 M

Q. $250 m L$ of sodium carbonate solution contains $2.65 g$ of $N a_{2} C O_{3}$ . If $10 m L$ of this solution is diluted to $500 m L$ , the concentration of the diluted acid will be

$0.01 M$

$0.001 M$

$0.05 M$

$0.002 M$

Molarity of $N a_{2} C O_{3}$ solution
$= \frac{2.65}{106} \times \frac{1000}{250} = 0.1 M$
$10 m L$ of this solution is diluted to $500 m L$
$M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times 10 = M_{2} \times 500$
$\Rightarrow M_{2} = 0.002 M$

Q. 250mL of sodium carbonate solution contains 2.65g of Na2​CO3​. If 10mL of this solution is diluted to 500mL, the concentration of the diluted acid will be

0.01M

0.001M

0.05M

0.002M