The correct option is B 0.2 L
For the neutralization reaction,
Na2CO3 + 2HCl→2NaCl + H2O + CO2
Given,
Specific gravity for Na2CO3 solution is 1.212 g/mL.
Let us assume that mass of the solvent be 1000 g.
Thus, moles of sodium carbonate =2 mol
Again, molar mass of Na2CO3=23×2+12+16×3=106 g/mol
mass of sodium carbonate will be =2×106 g=212 g
∴ mass of the solvent + solute
=1000+212=1212 g
Hence volume of the solution =massdensity=12121.212=1000 mL
molarity of sodium carbonate = moles volume=21=2 M
We know,
n-factor of sodium carbonate =2
n-factor of HCl=1
n-factor× molarity×volume of the base= n-factor× number of moles of the acid
2×2×0.025=1× number of moles of HCl
Hence, moles of HCl=0.1
mass of HCl= moles×molar mass
=0.1×36.5=3.65 g
Given, 1 litre of the HCl solution contains 18.25 g of HCl
volume of the solution needed =118.25×3.65=0.2 L