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Q.

250 mL of a ${Na}_{2} {CO}_{3}$ solution contains 2.65 g of ${Na}_{2} {CO}_{3}$ . 10 mL of this solution is added to X mL of water to obtain 0.001 M ${Na}_{2} {CO}_{3}$ solution. The value of X is:(Molecular mass of ${Na}_{2} {CO}_{3}$ = 106amu)

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a

990

b

1000

c

90

d

9990

answer is B.

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Detailed Solution

Molarity of solution

$M = \frac{w_{B} \times 1000}{m_{B} \times V} = \frac{2 .65 \times 1000}{106 \times 250} = 0 .1$

$M_{1} V_{1} = M_{2} V_{2}$

$0 .1 \times 10 = 0 .001 (10 + x)$

$x = 990 mL$

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250 mL of a Na2CO3 solution contains 2.65 g of Na2CO3. 10 mL of this solution is added to X mL of water to obtain 0.001 M Na2CO3 solution. The value of X is:(Molecular mass of Na2CO3 = 106amu)