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