An aqueous solution has a mole fraction of 0.072 KOH. What is the molarity of the solution? Assume that the density of the solution is 1.15 g/mL. Enter your answer to three significant figures and in units of molarity.

The molarity of the solution is 3.99 M

Explanation:

Step 1: Data given

Mole fraction = 0.072 KOH

Density of the solution = 1.15 g/mL

Step 2: Calculate mass of the solution

1000 mL * 1.15 g/mL = 1150 grams solution

Step 3: Calculate moles of KOH

moles KOH / (moles KOH + moles H2O) = 0.072

Suppose x = grams KOH

Moles of KOH = x/56g/mol

moles H2O = (1150 - x) / 18

1150 grams solution - grams KOH = grams of H2O

(x/56) / (x/56 + (1150 - x) / 18) = 0.072

0.017857x / (0.017857x + 63.89 - 0.0556x) = 0.072

x = 223.44 grams KOH

Moles of KOH = 223.44 grams / 56g/mol = 3.99 moles KOH

Step 4: Calculate molarity of the solution

223.44 grams KOH / 1 liter * 1 mole KOH / 56 g/mol = 3.99 mol/L

Molarity = 3.99 M