1.26 g of hydrated oxalic acid was dissolved in water to prepare 250ml of solution. calculate molarity of the solution.

Explanation

Volume of solution = 100mL
Density = 1.298 `"g mL"^(-1)`
Mass of solution = `100xx1.298=129.8g`
Mass of solute = `30/(100)xx129.8=38.94g`
Mass of solvent = 129.8-38.94=90.86 g
Volume of solution = 100 mL
Density = 1.610
Mass of solution = `100xx1.610=161g`
Mass of solute = `70/(100)xx161=112.7g`
Mass of solvent = 161 - 112.7 = 48.3g
Total mass of solute = 38.94 + 112.7 = 151.64g
Moles of solute = `151.64/(98)=1.55`
Mass of solvent = 90.86 + 48.3 = 139.16g
`"Molality"=(1.55xx1000)/(139.16)=11.1m,"Molarity"=(1.55xx1000)/(200)=7.75M`