Question

I need all of the following questions answered completely for chapter 13.

6) Calculate the freezing point of a water solution at each of these concentration.

part A) 0.75m

part B) 1.25m

part C) 4.5m

part D) 2.15m

7) A 140-mL sample of an 8.9 M NaCl solution is diluted to 2.0L.

What volume of the diluted solution contains 13.8g of NaCl?

8) A 157-mL sample of a 1.3M sucrose solution is diluted to 550mL.

what is the molarity of the diluted solution?

9) Which of the following mixtures are solutions?

Chose all that apply.

A) sand and water mixture

B) oil and water mixture

C) salt and water mixture

D) sterling silver cup

10) Ocean water contains 3.5% NaCl by mass.

What mass of ocean water in grams contains 50.8g of NaCl?

11) Calculate the concentration of each solution in mass percent.

Part A) 47.2g C12H22O11 in 468g H20

Part B) 140mg C6H12O6 in 4.91g H20

Part C) 2.00g NaCl in 175g H20

12) What volume of each solution contains 0.14 mol of KCl?

Part A) 0.287 M KCl

Part B) 1.9M KCl

Part C) 0.935 M KCl

13) To what volume should you dilute 35 mL of a 13 M stock HCl solution to obtain a 0.520 M HCl solution?

14) How many liters of a 0.600 M sucrose (C12H22O11) solution contain 1.6kg of sucrose?

15) Calculate the molarity of each solution.

Part A) 0.35 mol solute; 0.350 kg solvent

Part B) 0.832 mol solute; 0.250 kg solvent

Part C) 0.013 mol solute; 23.1 g solvent

for #15 it isnt molarity but molality**

Answer 1

6) The freezing point of each solution is calculated:

A) Tc = 0 - Kf * m = 0 - 1.86 * 0.75 = - 1.4 ° C

B) Tc = 0 - 1.86 * 1.25 = - 2.3 ° C

C) Tc = - 8.4 ° C

D) Tc = - 4.0 ° C

7) The final concentration is calculated:

C2 = C1 * V1 / V2 = 8.9 M * 140 mL / 2000 mL = 0.623 M

The volume of solution required is calculated:

V = 13.8 g NaCl * (1 mol / 58.44 g) * (1000 mL / 0.623 mol) = 380 mL

8) The final molarity is calculated:

C2 = 1.3 M * 157 mL / 550 mL = 0.371 M

9) Option C is a solution.