20 mL of a solution os sulphuric acid neuitralise 21.2 mL of 30% soution of sodium carbonete. How much water should be added to 100 mL of this solutio

Question

20 mL of a solution os sulphuric acid neuitralise 21.2 mL of 30% soution of sodium carbonete. How much water should be added to 100 mL of this solutio

20 mL of a solution os sulphuric acid neuitralise 21.2 mL of 30% soution of sodium carbonete. How much water should be added to 100 mL of this solutio to bring down its down its strength to decinoral?

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Calculation of normality of suplhric acid solution
strength of `Na_(2)CO_(3)` solution=`130g//100mL =300gL^(-1)`
` "Normality of" Na_(2)CO_(3) "solution"=("Strength")/("Equivalent mass")= (300gL^(-1))/(53g equiv^(-1))`
`=5.66"equiv." L^(-1)=5.66 N`.
`N_(1)xxV_(1)=N_(2)xxV_(2)or N_(1)=(N_(2)xxV_(2))/V_(1)`
`(H_(2)SO_(4))=(Na_(2)CO_(3))`
`N_(1)=((5.66N)xx(21.2mL))/((20.0mL))=6.0N`.
Calculation of volume of water to be added
Normality of `H_(2)SO_(4)` solution=6.0N
volume of `H_(2)SO_(4)` solution=100mL
Normality of diluted `H_(2)SO_(4)` solution=0.1N
Volume of diluted `H_(2)SO_(4)` solution may be calculated as:
`"underset"(("diluted" H_(2)SO_(4)))(N_(1)xxV_(1))=underset(("Concentreated"H_(2)SO_(4)))(N_(2) xx V_(2))`
`V_(1)=(N_(2)xxV_(2))/N_(1)=((6.0N)xx(100mL))/((0.1N))=6000mL`
Volume of water to be added =(6000-100)=5900mL