Question

A student dissolves 11. g of benzoic acid (C,H,02) in 500 ml of a solvent with a density of 1.19 g/ml. The student notices that the volume of the solve does not change when the benzoic acid dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity - O 0 0 0 molality - 0 x 5 ?

Answer 1

1)

Molar mass of C7H6O2,

MM = 7*MM(C) + 6*MM(H) + 2*MM(O)

= 7*12.01 + 6*1.008 + 2*16.0

= 122.118 g/mol

mass(C7H6O2)= 11 g

use:

number of mol of C7H6O2,

n = mass of C7H6O2/molar mass of C7H6O2

=(11 g)/(1.221*10^2 g/mol)

= 9.008*10^-2 mol

volume , V = 5*10^2 mL

= 0.5 L

use:

Molarity,

M = number of mol / volume in L

= 9.008*10^-2/0.5

= 0.1802 M

Answer: 0.180 M

2)

Mass of solvent = density * volume

= 1.19 g/mL * 500 mL

= 595 g

= 0.595 Kg

use:

Molality = number of mol / Mass of solvent in Kg

= 9.008*10^-2/0.595

= 0.1514 molal

Answer: 0.151 molal