Question

Calculate the molarity and mole fraction of 0.02 molal aqueous solution of naoh having Density 1.2g/ml

Answer 1

molality = 0.02

molality = moles of naoh / mass of solvent [ in Kg]

0.02 = moles of naoh / 1000g

moles of naoh = 20 mole

density = 1.2

density = mass of solution/ volume

1.2 = 1000/volume

volume = 1000/ 1.2

Molarity = moles ×1000/volume of solution

M = 20×1000/(1000/1.2)

= 20×1.2

= 24 M/mol

Mole fraction [ NaOH] = moles[ NaOH] / moles [ NaOH] + moles [ H2O]

= 20 / 20+( 1000/18)

= 0.26

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