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Q.

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)

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a

1000

b

990

c

9990

d

90

answer is B.

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Detailed Solution

Molarity of solution M=wB×1000mB×V=2.65×1000106×250=0.1M1V1=M2V20.1×10=0.001(10+x)x=990 mL
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