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 data: 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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.