Answer to Question #152137 in General Chemistry for yem

Question

A solution containing 46.0 g acetone, C3H6O, and 66.0 g of H2O, has a density of 0.926 g/mL. Calculate

a. The mass percentage of C3H6O

b. The mole fraction of C3H6O

c. The molality

d. The molarity of the solution

Answer 1

a.

Total mass = mass of acetone C3H6O+ mass of H2O

= (46+66)g

= 112g

mass percentage of C3H6O = (mass of acetone C3H6O×100)/Total mass %

= 41.07%

Hence, The mass percentage of C3H6O = 41.07%

b.

Molar Mass of C3H6O = 58 g/mol

So, no. Of Mole of C3H6O, n = (46/58)mol

Molar mass of H2O = 18 g/mol

So, no. Of moles of H2O, n'= (66/18) mol

So, mole fraction of C3H6O = n/(n+n')

= (46/58)/[(46/58)+(66/18)]

= 0.178

Hence, mole fraction of C3H6O = 0.178

c.

Total mass of solution, W1= 112 g

Mass of C3H6O, W2 = 46 g

Molar mass of C3H6O, M = 58 g/mol

Molality = W2×1000/(M×W1) mol/kg

= 46×1000/(58×112) mol/kg

= 7.08 mol/kg

Hence, molality = 7.08 mol/kg

d.

density of solution = 0.926 g/mL

Volume of the solution, V = (Mass of the solution/density of solution)

= (112/0.926)ml

= 120.95 ml

Mass of C3H6O, W2 = 46 g

Molar mass of C3H6O, M = 58 g/mol

Molarity = [W2×1000/(M×V)] M

= 6.557 M

Hence, molarity = 6.557 M