Question

30. What is the molarity when 17.26 grams of sodium sulfate is added to water to make a 1.50 L
solution?
31. What mass of water is required to dissolve 4.00 g of NaOH to prepare a 10.0% NaOH
solution?

Answer 1

we need molar mass of sodium sulfate, Na2SO4
= [2*22.99 + 1*32.06 + 4*16.0] g/mol = 142.04 g/mol

mass of Na2SO4 = 17.26 g
number of moles of Na2SO4 present = mass of Na2SO4/molar mass of sodium sulfate
= 17.26 g/142.04 g/mol = 0.1215 mols
volume of solution = 1.5 L

molarity = number of mols of solute/molar mass = 0.1215 mols/1.5 L = 0.081 M

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10 % NaOH means 10 g NaOH in 100 g solution

so amount of water = 100 g-10 g = 90 g
10 g NaOH in 90 g water
1 g NaOH in 90/10 =9 g water
4 g NaOH need 9*4 = 36 g of water
mass of water needed = 36g
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Thanks !