1.

Calculate the molarity and normality of solution containing 3.15 of hydrated oxalic acid (`{:(COOH),(|),(COOH):}.2H_(2)O`) dissolved in 250 mL of the solution.

Answer» Step I. Calculation of molarity of the sodium
Mass of oxalic acid `=3.15 g`
Molar mass of oxalic acid `= 24+64 +2 +2 xx 18 = 126 u = 126 "g mol"^(-1)`
No. of moles of oxalic acid `= ((3.15g))/((126"g mol"^(-1)))`
Volume of solution `= 250 mL=(250)/(1000)=0.25 L`
Molarity of solution `(M)=("No. of moles of oxalic acid")/("Volume of solution in litres")`
`= ((3.15g))/((126"g mol"^(-1))xx(0.25 L))`
`=0.1 "mol L"^(-1)=0.1 M`
Step II. Calculation of normality of the acid solution
Basicity of oxalic acid from the formula = 2
Equivalent mass of oxalic acid `= ("Molecular mass")/("Basicity")=((126 "g mol"^(-1)))/(2)=("63 g equiv"^(-1))`
No. of equivalent of oxalic acid `= ("Mass of oxalic acid")/("Equivalent mass of oxalic acid")=((3.15g))/(("63 g equiv"^(-1)))`
`=0.2 "equiv L"^(-1)=0.2 N`
Note : The normality of the acid solution can also be calculated with the help of relation :
Normality of acid = Molarity of acid `xx` Basicity `= 0.1 xx 2 = 0.2 N`


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