What is the molarity of a solution preparedby dissolving 10.0 g of acetone, C3H6O, in enough water to make 125 mL of solution?

1.379 M

Explanation:

The computation of the molarity of a solution is as follows;

As we know that

Molarity(M) = number of moles of solute ÷  volume of solution ( in L )

Given that

Mass = 10 grams

Molar mass of acetone(C3H6O) = 58 g/mol

moles = mass ÷ molar mass

= 10 g ÷ 58 g/mol

= 0.172414 mol

Now the molarity of the solution is  

Volume of solution = 125 mL = 0.125 L

As

1 L = 1000 mL

so,

125 mL = (125 mL  × ( 1 L ÷ 1000 mL) = 0.250 L)

Now the molarity is

= 0.172414 mol  ÷ 0.125 L

= 1.379 M