Question

250 ml of a sodium carbonate solution contains 2.65 gm of Na2CO3.10 ml of this solution is added to x ml of water to obtain 0.001M Na2CO3 solution. What is the value of x in ml?

Answer 1

990 ml is the answer.
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Answer 2

Answer:The value of x 990 mL.

Explanation:

Molarity of the solution of 250 ml of a sodium carbonate solution contains 2.65 gm of sodium carbonate .

$M_1=\frac{2.65 g}{105.98 g/mol\times 0.250 L}=0.1000 mol/L$

When 10 ml of above solution is added to x ml of water the molarity of the solution becomes 0.001 M

Volume of 0.1000 M solution = $V_1$

Molarity of the solution on dilution=$M_2$

Volume of the new solution = $V_2=10 ml+x ml$

$M_1\times V_1=M_2\times V_2$

$0.1000 M\times 10 ml=0.001 M\times (10 +x) mL$

On solving we get:

x = 990 mL

The value of x 990 mL.