Answer: Volume required is 0.115 L or 115 ml
Explanation:
moles of [tex]NaHCO_3[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{1.30g}{84g/mol}=0.015mol[/tex]
The balanced chemical equation is:
[tex]HCl+NaHCO_3\rightarrow NaCl+H_2CO_3[/tex]
1 mole of [tex]NaHCO_3[/tex] requires = 1 mole of HCl
Thus 0.015 mol of [tex]NaHCO_3[/tex] requires = [tex]\frac{1}{1}\times 0.015=0.015[/tex] mole of HCl
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in L
[tex]0.130=\frac{0..015}{V_s}[/tex]
[tex]V_s=0.115L[/tex]
Thus volume required is 0.115 l or 115 ml
