Calculate the molarity and mole fraction of acetone in a 1.17 m solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Molarity of acetone: 0.68 M

Mole fraction acetone: 1.17 / 18.37 = 0.063

Explanation:

First of all, think in this dа ta: 1.17 m. This is molal.

Molality means moles st / 1kg solvent so 1.17 moles are in 1 kg of ethanol.

Second step: Use Density to get volumen

Density = mass / volume

Ethanol density = Ethanol mass / ethanol volume

0.789 g/ml = 1000 g / volume ethanol

Volume ethanol = 1000g / 0.789 g/ml = 1267.4 ml

1.17, as we saw, are moles. A mass in g that we can obtained from the molar weight.

Molar weight acetone: 58.08 g/m

Moles. molar weight = mass → 58.08 g/mol. 1.17 mol = 67.9 g

Now let's use density

Acetone density = Acetone mass / Acetone volume

0.788 g/ml = 67.9 g / volume acetone

Volume acetone = 86.2 ml

Assume that the volumes of acetone and ethanol add

86.2 ml + 1267.4 ml = 1353.6 mL. - This is the volume of solution.

Molarity means moles of solute/L of solution. Let's think the rule of three:

1353.6 mL __ we have __ 1.17 moles

1000 mL ___ we have ___ (1000. 1.17) / 1353.6 = 0.86 M

For mole fraction, we need to know the moles of acetone.

Molar weight acetone 58.08 g/m

Mass / Molar weight : moles

1000 g / 58.08 g/m = 17.2 moles

Mole fraction acetone : acetone moles / total moles

Total moles: 17.2 + 1.17 = 18.37 moles

Mole fraction acetone: 1.17 / 18.37 = 0.063