Answer to Question #347623 in General Chemistry for Toniece

Question

Question 19

Suppose that a solution is prepared by mixing 25 mL (0.22 mol, 16 g) of pentane (the solute) with 45 mL (0.34 mol, 30 g) of hexane (the solvent). If the volumes add on mixing, answer questions 19-21 below.

 Calculate the molality of the pentane

a. 3.1 mol/kg

b. 4.8 mol/kg

c. 7.3 mol/kg

d. 3.1 mol/L

e. 7.3 mol/L

Question 20

Calculate the molarity of the pentane?

a. 3.1 mol/kg

b. 7.3 mol/kg

c. 3.1 mol/L

d. 7.3 mol/L

e. 45.8 mol/L

Question 21

What is the mole fraction of the pentane?

a. 0.35

b. 0.39

c. 0.64

d. 39%

e. 64 %

 Question 22

If I prepare a solution by adding 25 mL of 2.0 M NaOH into a 500 mL volumetric flask, and filling the flask to the mark with water, what will the final concentration of the solution be?

a. 0.05 M

b. 0.10 M

c. 0.5 M

d. 2.0 M

e. 40.0 M

Answer 1

19.

C_molality = mol of solute / kg of solvent

C_molality = 0,22 mol / 0,03 kg = 7,3 mol/kg

Answer: C

20.

C_M = n / V;

C_M(pentane) = 0,22mol / 0,07 l = 3,1 mol/l

Answer: C

21.

Mole fraction_(pentane) = n_(pentane) / n_(total)

Mole fraction_(pentane) =0,22/(0,22+0,34)= =0,39

Answer: B

22.

n(NaOH) = C_M * V=2mol/l *0,025 l =0,05mol.

for the post-dilution:

C_M = n/V = 0,05mol / 0,5 l = 0,1 mol/l

Answer: B