Question

The volune of 0.025 m ca(oh)2 solution qhich can neutralise 10.6 g of anhydrous na2co3 is

Answer 1

Heya friends.

This is the given reaction according to your question :-

3 Ca(OH)₂ + Na₂CO₃ → Na(OH)₂ + 6 CaCO₃ 
moles of Na₂CO₃ can be calculated
= given mass / Molar mass (which is 10.6 g) 

Hence, moles of Sodium carbonate
= 10.6 / 10.6 = 0.1
Here for 0.1 moles we need 0.2 moles of HCl.
$Now, Molarity = moles of HCl / Volume of HCl$
= 2m
Here it is equal to 2m 


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