Question

The molarity of 200 ml of HCl solution which can neutralise 10.6 g. of anhydrous Na2CO3 is
1. 0.1M
2. 1M
3.0.6M
4. 0.75M​

Answer 1

The equation for the neutralization is as shown:

Na2CO3 + 2HCl--------> 2NaCl + CO2 + H2O

So, 1 part of Na2CO3 neutralizes 2 parts of HCl.

molar mass of Na2CO3 is 2 X 23 + 12 + 3X16 => 46 + 12 +48 => 106 g mol-1

So, 106 X 0.1 grams of Na2CO3 is 1 X 0.1 => 0.1 moles of Na2CO3.

Hence, from the equation, 1 part of Na2CO3 is 0.1 moles.

So, 2 parts of HCl is 2 X 0.1 => 0.2 moles of HCl gas.

200 ml of solution should contain 0.2 moles of HCl gas, molarity=?

200 ml is 0.2 L

molarity = (moles of HCl)/litres of solution => 0.2/0.2 => 1 M

So, the molarity of the solution will be 2) 1 M