KVPY2019ChemistrySolutions
3 . 0   g of oxalic acid CO 2 H 2 · 2 H 2 O is dissolved in a solvent to prepare a 250   mL solution. The density of the solution is 1 . 9   g / mL . The molality and normality of the solution, respectively, are closest to
Options
- A0 . 10   mol   kg - 1 and 0 . 38   N
- B0 . 10   mol   kg - 1 and 0 . 19   N
- C0 . 05   mol   kg - 1 and 0 . 19   N
- D0 . 05   mol   kg - 1 and 0 . 09   N
Correct answer
C. 0 . 05   mol   kg - 1 and 0 . 19   N
Step-by-step solution
Molality = Mass of solute × 1000 Molar mass of solute × Mass of solvent Mass of solvent = 250 mL × 19 g / mL = 475 g ∴ Mass of solvent = 472 g Molar mass of CO 2 H 2 . 2 H 2 O = 126 g mol - 1 ∴ Molality = 3 × 1000 472 × 126 = 0 . 05 mol kg - 1 Normality = Number of equivalents of solute Volume of solution l = Mass of solute g × 1000 Equivalent mass of solute × Volume of solution mL Equivalent mass of oxalic acid = Molar mass 2 = 63 g / equi . ∴ Normality = 3 × 1000 63 × 250 = 0 . 19 (equivalents) / l or 0 . 19 N