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Q.

The molarity of 200ml of HCl  solution which can neutralize 10.6g of anhydrous Na2CO3 is ____?

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a

0.1M

b

1M

c

0.6M

d

0.75M

answer is B.

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Detailed Solution

milli equivalents of acid = milli equivalents of baseNV1000HCl=wENa2CO3N×2001000=10.6×2106N=0.2×1000200N = 1Normality = molarity × basicity of acid 1 = M ×1M = 1M
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The molarity of 200ml of HCl  solution which can neutralize 10.6g of anhydrous Na2CO3 is ____?