Answer to Question #132961 in Physical Chemistry for Mike

Question

Complete required calculations and shortly describe how you are going to make showed solutions from already given products?
0.0250L 0.0150mol/L Na2SO4 solution from Na2SO4*10H2O(s)
250g 5% Na2SO4 solution from Na2SO4*10H2O(s)

Answer 1

1) First, the number of moles of Na₂SO₄ in the needed solution should be determined:

n (Na₂SO₄) = c*V = 0.0150mol/L * 0.0250L = 0.000375 mol

1 mole of Na₂SO₄*10H₂O contains one mole of Na₂SO₄, hence to get 0.000375 moles of Na₂SO₄, 0.000375 moles of Na₂SO₄*10H₂O should be used. Now, the mass of the hydrate should be calculated:

m (Na₂SO₄*10H₂O) = 0.000375mol * 322.195g/mol = 0.121 g

Therefore, to prepare the required solution, 0.121 grams of Na₂SO₄*10H₂O should be weighted and added to a 25 mL volumetric flask. Then the distilled water should be added to the same flask while shaking the mixture gently until the volume reaches 0.0250 L (or 25.0 mL) mark.

2) The mass of Na₂SO₄ in the solution is:

250g * 0.05 = 12.5 g.

To find the mass of Na₂SO₄*10H₂O required to produce 12.5 grams of Na₂SO₄, the mass percent of Na₂SO₄ in the hydrate should be determined:

% Na₂SO₄ = (M(Na₂SO₄) / M(Na₂SO₄*10H₂O)) * 100% = (142.04 / 322.195) * 100% = 44.085%.

m (Na₂SO₄*10H₂O) = 12.5g / 0.44085 = 28.4 g

To prepare the required solution, 28.4 grams of Na₂SO₄*10H₂O are needed. Therefore, the mass of water is 250 - 28.4 = 221.6 g. Assuming the density of pure water equal to 0.998 g/mL at 20^oC, the required volume of water is 221.6g / 0.998g/mL: = 222.0 mL.

Therefore, to prepare the required solution, 28.4 grams of Na₂SO₄*10H₂O should be weighted and added to the glassware with 222.0 mL of pure water, and the mixture should be stirred.