PERCENT BY MASS

1a. How many moles of solute are contained in 343 grams of a 23% aqueous solution of MgCr₂O₇?

1b.How many grams of solute are contained in the solution of question 1a?

1c. How many grams of water (the solvent) are contained in the solution of question 1a?

1d. How many molecules of water are contained in the solution of question 1a?

1e. How many Cr atoms are contained in the solution of question 1a?

2a. The density of a 0.13% solution of NaCl is 2.16 g/mL. What mass of NaCl would be required to prepare 45L of this solution?

2b. How many molecules of solute are contained in the solution of question 2a?

3a. Specific gravity of a 33% (NH₄)₂SO₄ solution is 1.1. What mass of (NH₄)₂SO₄ would be required to prepare 201mL of this solution?

3b. How many moles of solute are contained in the solution of question 3a?

ANSWERS

1a) ? g MgCr₂O₇ = 343g x 0.23 = 78.89 g MgCr₂O₇

? mol MgCr₂O₇ = 78.89 g MgCr₂O₇ x (1mol / 240g MgCr₂O₇ ) = 0.329 mol MgCr₂O₇

1b) ?g MgCr₂O₇ = 343g x 0.23 = 78.89 g MgCr₂O₇

1c) ? g H₂O = 343 g solution – 78.89g MgCr₂O₇ = 264.11 g H₂O

1d) ? moles H₂O = 264.11 g H₂O x (1 mol/18 g H₂O)=14.67 mol H₂O

? Molecules H₂O= 14.67mol H₂O x (6.02 x 10²³molecules/1mol H₂O)= 8.83 x 10²⁴ molecules H₂O

1e) ? molecules of MgCr₂O₇ = 0.329 mol MgCr₂O₇ x (6.02 x 10²³ molecules/ 1 mol)

= 1.98 x 10²³ molecules of MgCr₂O₇

? atoms of Cr = 1.98 x 10²³ molecules of MgCr₂O₇ x ( 2 Cr atoms/ 1 molecule) = 3.96 x 10²³ Cr atoms

2a) ? g NaCl impure = (2.16 g NaCl /1 mL) x (1000 mL / 1 L) = 2160 g/L impure NaCl

?g NaCl pure= (2160 g NaCl/ L) x 0.0013 = 2.808 g / L NaCl pure

? g NaCl = 45 L x (2.808 g NaCl/ 1 L) = 126.36 g NaCl

2b) ? moles NaCl = 126.36 g NaCl x (1mol / 58g NaCl)= 2.17 mol NaCl

? molecules NaCl = 2.17 mol NaCl x (6.02 x 10²³ molecules/ 1 mol) = 1.31 x 10²⁴ molecules NaCl

3a) specific gravity = density

? g (NH₄)₂SO₄ impure = (1.1g (NH₄)₂SO₄/ 1 mL) x (1000mL / 1L) = 1100g (NH₄)₂SO₄/ L impure

? g (NH₄)₂SO₄ pure= (1100 g (NH₄)₂SO₄ / L)(0.33) = 363 g (NH₄)₂SO₄ /L pure

? g (NH₄)₂SO₄ = 0.201L x (363g (NH₄)₂SO₄ / L) = 72.96 g (NH₄)₂SO₄

3b) ? moles (NH₄)₂SO₄ = 72.96 g (NH₄)₂SO₄ x (1 mol/ 132g (NH₄)₂SO₄) = 0.55 mol (NH₄)₂SO₄