Question

1a. If 20.0 grams of sucrose (C12H22O11) were diluted to 100.0 mL with water, calculate:

- Mass percent

- Molarity

1b. If 42.5 mL of isopropanol (C3H7OH) were added to 95.0 mL of water, calculate:

- Volume percent

- Molarity (the density of isopropanol is 0.786 g/mL)

1c. If 65.7 grams of BaCl2 were dissolved in water, and the total solution volume was 145.0 mL, calculate:

- Molarity of BaCl2

- Molarity of Cl-

Answer 1

MOlarity of solution = moles/ V(L) = [mass/molar mass]/V(l)

mass percent = mass of solute x100/ mass of solution

1a. If 20.0 grams of sucrose (C12H22O11) were diluted to 100.0 mL with water, calculate:

- Mass percent

- Molarity

mass of solute = 20.0g

molar mass of sucrose = 342g/mol

volume of solution = 100.0mL = 0.10 L

MOlarity = [ 20g/342g/mol] /0.1 0L

= 0.5847 M

to calculate the Mass percent we need either mass of solvent used (volume of wate used) or density of solution.

1b. If 42.5 mL of isopropanol (C3H7OH) were added to 95.0 mL of water, calculate:

- Volume percent

- Molarity (the density of isopropanol is 0.786 g/mL)

% volume = volume of solute x100 / total volume of solution

= 42.5mL x100 / (42.5+95) mL

= 30.91 %

mass of solute = volume x density = 42.5mL x 0.786g/ml =33.405 g

Thus

molarity = [33.405g/60g/mol] /0.095L =5.86 M

1c. If 65.7 grams of BaCl2 were dissolved in water, and the total solution volume was 145.0 mL, calculate:

- Molarity of BaCl2

- Molarity of Cl-

molarity of BaCL2 = [65.7g /208.23g/mol] /0.145L

= 2.174 M

molarity of Cl- in solution = 2 x molarity of BaCL2 [ as one mole BaCl2 gives 2 Cl- in solution]

= 2x 2.174 = 4.35 M