Chemistry

Answers to Chapter 15 Study Questions

 

1.

mass KMnO4 = 1.00 mole = 158 g; mass solution = 158 g KMnO4 + 158 g H2O = 316 g

= 50.0 %

2. = 0.466 moles

3. = 0.138 moles

4. = 29.2 g

5. V1 x M1 = V2 x M2; V1 x 2.00 M = 125 mL x 0.350 M

V1 = 125 mL x 0.350 M/2.00 M = 21.9 mL

6. = 2.50 M

= 0.772 M

8. a) CaCl2(aq) + 2 AgNO3(aq) 2 AgCl(s) + Ca(NO3)2(aq)

b) = 24.0 mL

9. VA x MA = VB x MB; VB x 0.400 M = 16.0 mL x 0.120 M

VB = 16.0 mL x 0.120 M/0.400 M = 4.80 mL

10. Colligative properties: vapor pressure, boiling point, freezing point.

Not colligative properties: color, electrical conductivity, density

11. Highest boiling point = greatest number of solute particles.

c) 1.0 M Ca(NO3)2 > d) 1.0 M MgSO4 > b) 1.0 M glucose (C6H12O6) > a) pure water

3 mol particles/mol 2 mol particles/mol 1 mol particles/mol

12. a) DTf = 1.86 C x moles solute particles/kg water

= 3.72C; Tf = 0 - DTf = 0 - 3.72C = -3.72C

b) DTf = 1.86 C x moles solute particles/kg water

= 0.300 moles

c) DTf = 1.86 C x moles solute particles/kg water

= 0.751 mol particles

= 5.59C; Tf = -5.59C

(CaCl2 is an electrolyte and it is important to remember that there are 3 moles of ions per mole of CaCl2. The freezing point is three times lower than it would be for a nonelectrolyte.)