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Molarity

250 mL of a ...

Question

250 mL of a $N a_{2} C O_{3}$ solution contains 2.65 g of $N a_{2} C O_{3} .$ 10 mL of this solution is added to x mL of water to obtain 0.001 M $N a_{2} C O_{3}$ solution. The value of x is ______.
(Molecular mass of $N a_{2} C O_{3} = 106 a m u$ )

A

1000

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B

990

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C

9990

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D

90

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Solution

The correct option is A 990
Molarity of solution.
$M = \frac{w_{B} \times 1000}{m_{B} \times V} = \frac{2.65 \times 1000}{106 \times 250} = 0.1$
$M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times 10 = 0.001 (10 + x)$
$x = 990 m L$

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