A solution of oxalic acid, `(COOH)_(2).2H_(2)O` is prepared by dissolving 0.63 g of the acid in 250 `cm^(3)` of the solution. Calculate molarity

Explanation

Calculation of molarity.
`"Molar mass of oxalic acid, " {:(" "COOH),(" |"),(" "COOH):}.2H_(2)O=126gmol^(-1)`
0.63 g of oxalic acid `=(0.63)/(126)` mole of oxalic acid = 0.005 mole of oxalic acid `["Moles"=("Mass in g")/("Molar mass")]`
Volume of the solution `=250cm^(3)=0.250L`
`"Molarity of the solution"=("Moles of the solute")/("Volume of sol. in L")=(0.005 mol)/(0.250L)=0.02molL^(-1)`
i.e., Molarity of the solution = 0.02 M