A student dissolves 13. g of stilbene (C14H12) in 100. mL of a solvent with a density of 0.85 g/mL. The student notices that the volume of the solvent does not change when the stilbene dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

Molarity = 0.72M

Molality = 0.85M

Explanation:

Molarity and molality are two measures of molar concentration, and they can be calculated as follows:

Molarity = number of moles/volume of solvent

Molality = number of moles/mass (kg) of solvent

Mole = mass/molar mass

Molar mass of stilbene (C14H12) = 12(14) + 1(12)

= 168 + 12

= 180g/mol

mole of stilbene = 13/180

mole = 0.072mol

Volume = 100mL = 0.1L

molarity = 0.072/0.1

Molarity = 0.72M

Molality = number of moles ÷ mass (kg) of solvent

To get mass of solvent, we use;

Density = mass/volume

Mass = density × volume

Mass = 0.85g/mL × 100mL

Mass = 85g

Mass in kg = 85/1000 = 0.085kg

Molality = 0.072/0.085

Molality = 0.847M

Molality = 0.85M