Question

What volume of a 50.0% by mass NaOH solution must be diluted to prepare a 500 mL of 0.3500 M NaOH solution? The density of 50% NaOH solution is 1.515 g/ml at 25 degrees C.

Answer #1

Volume of solution after dilution, Vf = 500 mL = 500 mLx(1L / 1000mL) = 0.500 L

Concentration after dilution, Mf = 0.350 M

Hence moles of NaOH in the solution, n = MfxVf = 0.350x 0.500 = 0.175 M

Since the number of moles of solute remains constant before and after dilution,

Composition before dilution is 50% by mass NaOH solution.

Let the mass of 50 % NaOH solution be 'm' g

mass of NaOH in the solution = m x (50/100) = m/2 g

and mass of water in the solution = m - m/2 = m/2 g

Hence moles of NaOH in the solution before dilution = mass of NaOH / molar mass of NaOH

= (m/2) g/ 40gmol^{-1} = m / 80 mol

Since the number of moles of solute remains constant before and after dilution, hence

m/ 80 mol NaOH = 0.175 M NaOH

=> m = 0.175x80 = 14 g

Hence mass of NaOH = m/ 2 = 7 g and mass of water = m/2 = 7 g (before dilution)

Hence volume of NaOH solution before dilution,Vi = m / d = 14 g
/ 1.515 g/ml = 9.24 mL = 9.24x10^{-3} L

Concentration of NaOH solution before dilution, Mi = moles of
NaOH / volume(L) = (7/40) / 9.24x10^{-3} L

= 18.94M

Hence the volume added to 9.24 mL 18.94 M NaOH solution to prepare 500 mL of 0.3500 M NaOH solution

= 500 - 9.24 = 490.76 mL (answer)

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