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Question

25 mL of a solution of 2 m Na2CO3 having a specific gravity of 1.212 g/mL requires what volume of a solution of HCl, containing 18.25 g of the acid per litre, for complete neutralization?

A
0.05 L
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B
0.2 L
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C
0.5 L
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D
1 L
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Solution

The correct option is B 0.2 L
For the neutralization reaction,
Na2CO3 + 2HCl2NaCl + 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

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