Answer to Question #154785 in Chemistry for Yaprak

Question

What is the density of 54.3 % (wt/wt) aqueous NaOH (MW = 40.00) if 17.6 mL of the solution diluted to 2.00 L gives 0.196 M NaOH?

Answer 1

M₁V₁ = M₂V₂

M₁ = molarity of conc. NaOH (unknown)

V₁ = volume of conc. NaOH

M₂ = molarity of diluted NaOH

V₂ = volume of diluted NaOH

M₁ = (M₂V₂ / V₁) = (0.169)(2000 mL) / (16.7 mL) = 20.2 M

20.2 moles NaOH / L x (40.0 g NaOH / 1 mole NaOH) = 808 g NaOH / L

53.4 mass % NaOH = 53.4 g NaOH / 100 g solution

How much solution contains 808 g NaOH?

(53.4 g NaOH / 100 g soln.) = (808 g NaOH / x g solution). By proportion, x = 1510 g

So 1510 g = 1L

D = g/mL = 1510 g / 1000 mL = 1.510 g/mL