Question

a solution of oxalic acid is prepared by dissolving0.63g of the acid in 250 cm cube of solution . calculate molarity

Answer 1

Moles of solute=mass given/molar mass
0.63/126=0.005m

Now change the volume into liters
250/1000=0.25l
M=moles of solute/vol of solution(in liters)
0.005/0.25=0.25M

Answer 2

Given conditions ⇒   Volume of the Solution = 250 cm³.  = 250 mL. = 250 × 10⁻³ L.   Mass of the Oxalic Acids = 0.63 grams.  Molar Mass of the Oxalic Acid = 1236 g/mole.   For molarity,  ∵ No. of moles = Mass/Molar Mass ∴ No. of moles = 0.63/126     = 5 × 10⁻³ moles.   Using the Formula,   Molarity (M)  = No. of moles of the Solute/Volume of the Solution in liter  ∴ M = (5 × 10⁻³)/(250 × 10⁻³)    M = 0.02 M  Hence, the molarity of the solution is 0.02 M.   For Normality,   Basicity of the Oxalic Acid = 2 Normality = Basicity x Molarity  = 2 x 0.02 = 0.04 N.    For Molality,   Molarity calculated = 0.02 M. This means that 0.02 moles of the  acid is present in the 1 liter of the solution.   Now, Assuming that the Total number of moles in 1 Liter of the solution is 1.   No. of moles of water = 1 - 0.002 moles.  = 0.98 moles  Therefore, mass of the water = 0.98 x 18 = 17.64 g.   Now,  Molality(m) = No. of moles of solute/Mass of the solvent or water = (0.005/17.64) x 1000 = 0.28 m.   Hence, molality of the solution is 0.28 m Please mark me as brainliest