A 100 mL solution that is 0.02 M in Na+ can be prepared by adding how many g of Na2SO4 (of MW 142 g/mol)?

Question

to water and diluting a final volume of 100 mL?

1. 23.8 g
2. 0.028 g
3. 0.045 g
4. 0.28 g
5. 0.14 g

Answer 1

Let the solution be called S

100mLS x 0.02molNa/1000mLS x 1molNa2SO4/2molNa x 142gNa2SO4/1molNa2SO4 = 0.14g Na2SO4

5. Is correct

Answer 2

0.02 moles Na^+1/L x 1L/1000 mL x 100 mL = 0.002 moles Na

0.002 moles Na^+2 x 1 mole Na2SO4/2 mole Na x 142 grams/mole Na2SO4 = 0.142 grams Na2SO4

5

Answer 3

Answer should be 0.14 grams