Question

1- Express the concentration of a 0.0420 M0.0420 M aqueous solution of fluoride, F−,F−, in mass percentage and in parts per million (ppm). Assume the density of the solution is 1.00 g/mL.1.00 g/mL.

mass percentage:

ppm:

2- A solution is made by dissolving 0.618 mol0.618 mol of nonelectrolyte solute in 795 g795 g of benzene. Calculate the freezing point, Tf,Tf, and boiling point, Tb,Tb, of the solution. Constants can be found in the table of colligative constants.

Tf=

Tb=

Solvent	Formula	Kf value* (°C/m)	Normal freezing point (°C)	Kb value (°C/m)	Normal boiling point (°C)
water	H2O	1.86	0.00	0.512	100.00
benzene	C6H6	5.12	5.49	2.53	80.1
cyclohexane	C6H12	20.8	6.59	2.92	80.7
ethanol	C2H6O	1.99	–117.3	1.22	78.4
carbon tetrachloride	CCl4	29.8	–22.9	5.03	76.8
camphor	C10H16O	37.8	176

Answer 1

Solution:

A) Calculation of mass percent:

Molarity (M) is calculated as,

M = Mass / Molar mass x Volume in L

Molar mass of F- = 8.998 g mol-1

Therefore,

0.0420 mol /L  = Mass of F- / 18.998 g mol-1 x 1L

>>Mass of F- = 0.0420 x 18.998 g = 0.7979 g

Mass percent = (Mass of F- / Mass of solution ) x 100

Since, density = 1g / mL

Hence, Mass of solution = Volume of solution

1 L = 1 Kg = 1000 g

Thus,

Mass percent = (Mass of F- / Mass of solution ) x 100 = (0.7979 g / 1000 g) x 100

Mass percent = 0.07979 g = 0.0798 %

B) Calculation of ppm :

ppm is calculated as,

ppm = 10000 x % Mass = 1000 x 0.07979 = 797.9 ppm