1.

6.3 g of exalic acid (eq.wt=63) is dissoved in 500 mL. solution. What will be the normality of the solution ?

Answer»


Solution :`"Normalioty of solution (N)"=("Mass of oxalic acid/EQUIVALENT mass")/("VOLUME of solution in litres")`
`"Mass os oxalic acid = 6.3 g, Equivalent mass of oxalic acid "= 63" g equiv"^(-1)`,
`"Volume of solution=500 mL="500/1000=0.5 L`
`"Normality (N)"=((6.3g)/(63" g equiv"^(-1)))/((0.5L))=0.2"equiv L"^(-1)=0.2N.`


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