Question

Aqueous solution of naoh is marked 10 percent w/ w the density of the solution is 1.070 g/ cm3 calculate molarity molality mol fraction of naoh in water

Answer 1

V=m/d

molarity=10×(1.07/4+90/18)

=1.89

Answer 2

Answer:

Molarity = 2.675 M

Molality = 2.777 m

Mole fraction (NaOH) = 0.0476  

Mole fraction (H2O) = 0.9523

Explanation:

aqueous solution of NaOH is marked 10%(w/w). Dwnsity of solution is 1.070 gm/cm^3.

(i) Molarity -

Molarity = no of moles of solute / volume of soln

Molarity = (10/40) / (0.1÷1.07)

Molarity = 2.675 M

(ii) Molality -

Molality = no of moles of solute / mass of solvent

Molality = (10/40) / (0.090)

Molality = 2.777 m

(iii) Mole fraction of NaOH -

Mole fraction (NaOH) = moles of NaOH / total moles

Mole fraction (NaOH) = (10/40) / (10/40 + 90/18)

Mole fraction (NaOH) = 0.0476

(iv) Mole fraction of H2O -

Mole fraction (H2O) = moles of H2O / total moles

Mole fraction (H2O) = (90/18) / (10/40 + 90/18)

Mole fraction (H2O) = 0.9523