The molarity of the HCl solution : 1.052 M
Further explanation
Given
Volume of HCl = 38.55 ml
mass of Na₂CO₃ (MW=106 g/mol)=2.15 g
Reaction
Na₂CO₃(aq) + 2HCl (aq)→ 2NaCl (aq) + CO₂ (g) + H₂0 (l)
Required
The molarity of the HCl solution
Solution
mol Na₂CO₃ :
[tex]\tt mol=\dfrac{mass}{MW}=\dfrac{2.15}{106}=0.02028[/tex]
From the equation, mol ratio Na₂CO₃ : HCl = 1 : 2, so mol HCl :
[tex]\tt mol~HCl=2\times 0.02028=0.04056[/tex]
The molarity :
[tex]\tt M=\dfrac{n}{V}=\dfrac{0.04056}{0.03855~L}=1.052~M[/tex]
