250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. wt. of Na2CO3 = 106)

Question

250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. wt. of Na2CO3 = 106) (A) 0.1 M (B) 0.001 M (C) 0.01 M (D) 10-4 M

Tardigrade Admin · Accepted Answer

Molarity = (W(g) × 1000/ textmolecular wt. × V(ml.)) = (2.65 × 1000/106 × 250) = 0.1 M 10 ml of this solution is diluted to 1000 ml N1 V1 = N2V2 10 × 0.1 = 1000 × x x = (0.1 × 10/1000) = 0.001 M

Q. $250 m l$ of a sodium carbonate solution contains $2.65$ grams of $N a_{2} C O_{3}$ . If $10 m l$ of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. wt. of $N a_{2} C O_{3} = 106)$

$0.1 M$

$0.001 M$

$0.01 M$

$1 0^{- 4} M$

Molarity $= \frac{W ( g ) \times 1000}{molecular wt. \times V ( m l . )}$
$= \frac{2.65 \times 1000}{106 \times 250} = 0.1 M$
$10$ ml of this solution is diluted to $1000$ ml $N_{1} V_{1} = N_{2} V_{2}$
$10 \times 0.1 = 1000 \times x$
$x = \frac{0.1 \times 10}{1000} = 0.001 M$

Q. 250ml of a sodium carbonate solution contains 2.65 grams of Na2​CO3​. If 10ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. wt. of Na2​CO3​=106)

0.1M

0.001M

0.01M

10−4M