Question

A one-liter bottle of a typical sports drink contains 426 mg of sodium ions, 113 mg of potassium ions, and 53 grams of sugars, and provides 190 calories. If sodium chloride is the source of the sodium ions, potassium hydrogen phosphate (K2HPO4) is the source of the potassium ions, and sucrose (C12H22O11) is the only sugar.

PART A:

Calculate the molarity of sodium chloride in the bottle. (Hint: One mole of sodium ion is contained in one mole of sodium chloride.)

Express your answer to three significant figures.

PART B:

Calculate the molarity of potassium hydrogen phosphate in the bottle. (Hint: Two moles of potassium ions are in one mole of potassium hydrogen phosphate.)

Express your answer to three significant figures.

PART C:

Calculate the molarity of sucrose in the bottle.

Express your answer to two significant figures.

Answer 1

A) Mass of sodium ions in the bottle = 426 mg.

Volume of the bottle = 1 L.

The atomic masses are

Na: 22.989 g/mol

Mol(s) sodium ions corresponding to 426 mg = (426 mg)/(22.989 g/mol)

= (426 mg)*(1 g)/(1000 mg)/(22.989 g/mol)

= 0.01853 mol = mols sodium chloride (since 1 mol sodium chloride gives 1 mol of sodium ions).

Molarity of sodium chloride in the bottle = (mols sodium ion)/(volume of the bottle)

= (0.01853 mol)/(1 L)

= 0.01853 mol.L-1

≈ 0.0185 mol.L-1 (ans, correct to 3 sig. figs).

B) Mass of potassium ions in the bottle = 113 mg.

The atomic masses are

K: 39.098 g/mol

Mol(s) potassium ions corresponding to 113 mg = (113 mg)/(39.098 g/mol)

= (113 mg)*(1 g)/(1000 mg)/(39.098 g/mol)

= 0.002890 mol

Mol(s) K2HPO4 = (0.002890 mol potassium ions)*(1 mol K2HPO4)/(2 mols potassium ions)

= 0.001445 mol (1 mol K2HPO4 gives 2 mols of potassium ions).

Molarity of K2HPO4 in the bottle = (mols potassium ion)/(volume of the bottle)

= (0.001445 mol)/(1 L)

= 0.001445 mol.L-1

≈ 0.00144 mol.L-1 (ans, correct to 3 sig. figs).

C) Mass of sucrose in the bottle = 53 g

The gram molar masses are

C: 12.011 g/mol

H: 1.008 g/mol

O: 15.999 g/mol

Gram molar mass of sucrose = (12*12.011 + 22*1.008 + 11*15.999) g/mol

= 342.297 g/mol

Mol(s) sucrose corresponding to 53 g = (53 g)/(342.297 g/mol) = 0.1548 mol.

Molarity of sucrose in the bottle = (mols sucrose)/(volume of the bottle)

= (0.1548 mol)/(1 L)

= 0.1548 mol.L-1

≈ 0.15 mol.L-1 (ans, correct to 2 sig. figs).