Question

the molarity of 200ml of hcl solution which can neutralize 10.6g of anhydrous na2co3​

Answer 1

Given: wt. of hcl = 200ml

wt. of na2co3 = 10.6g

Molecular wt. of na2co3 = 106

Molarity = no. of moles of solute/ Volume of solution in liters.

n= 0.1 by calculation. n=given wt./MW

M = 0.1/ 200

= 0.0005 n/ml

we need it in Liters

therefore M = 0.0005* 10^3

Molarity = 0.5 M

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