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

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