Problems 

1. What is the molality and % mass if 1500 ml of water has 116 g of Potassium Fluoride dissolved in it.

•         Molality=moles solute/Kg solvent

•         Potassium Fluoride  =KF

•         K is 39 g/mol+F is 19g/mol=58 g/mol

•         116/58= 2 moles

•         1500 ml (1g/ml)=1500 g

•         1500 g/1000=1.5 kg

•         2moles/1.5 Kg=1.33 m

2.What is the molality and % mass if 1500 ml of water has 116g of potassium Fluoride added to it.

•         % mass=mass solute/Mass solution

•         1500(1g/ml)=1500 g

•         Solute + solvent=solution

•         1500 g +116 g =1616 g

•         116/1616 X 100 %

•         7.2 %

3.What is the molality and % mass if 2000 ml of alcohol (density =.8 g/ml) has 116g of potassium Fluoride dissolved in it.

•         Molality=moles solute/Kg solvent

•         39 + 19=58 g/mol

•         116 g/58 g/mol= 2moles

•         2000 ml(.8 g/ml)=1600 g

•         1600 g/1000=1.6 kg

•         2 moles/1.6 Kg=1.25 m

4. What is the % mass if 2000 ml of alcohol(.8 density) has 116g of potassium Fluoride dissolved in it.

•         % mass=mass solute/Mass solution

•         2000 ml(.8 g/ml)=1600 g

•         1600 +116=1716

•         116/1716(100)

•         6.8 %

5. If 300 ml of H3PO4 (2 moles) is added to 1200ml of water, calculate % volume.

•         % volume=vol. solute/vol solution

•         Solution = solute + solvent

•         Solution = 300 ml + 1200 ml= 1500 ml

•         300 ml/1500 ml (100)= 20 %

6. If 300 ml of H3PO4 (2mole) is added to 1200ml of water, calculate Molarity.

•         Solution=300 + 1200=1500ml

•         Molarity =mol solute/L solution

•         1500 ml/1000=1.5 L

•         2 mole/1.5L= 1.3 M

7. If 300 ml of H3PO4 (2 mole) is added to 1200ml of water,  calculate Normality.

•         Normality= equiv/Liters solution

•         H3= 3 H+1 ions

•         3 H+1 ions(2 moles) = 6 equiv

•         1500ml/1000=1.5 L

•         6 equiv./1.5 L =  4 N

8. What is the molality and % mass if 3000 ml of alcohol(.8 g/ml density) has 267g of Aluminum Bromide dissolved in it. Calculate the new boiling point.

•         Molality=moles solute/Kg solvent

•         Al=27 Br=80  27 + 3(80)=267g/mol

•         267 g/267 g/mol= 1 mole

•         3000 ml(.8 g/ml)=2400 g

•         2400 g/1000=2.4 kg

•         1moles/2.4 Kg =0.42 m

9. What is the molality and % mass  if 3000 ml of alcohol(.8 g/ml density) has 267 g of Aluminum Bromide dissolved in it.

•         % mass=mass solute/Mass solution

•         3000 ml(.8 g/ml)=2400 g

•         Solvent + solute = solution

•         2400g of alcohol +267g of AlBr3= 2667 g

•         267 g/2667 g

•         10 %

•         Boiling point=78 oC

•         molal ( Kb)

•         .42 m(1.19 oC/m)=0.4998 oC

•         Boiling point + boiling point elevation

•         78 oC + .4998 oC=78.4998 oC

10. If 1000 ml of H3PO4 (5 moles) is added to 2000ml of water, calculate  % volume.

•         % volume=vol.solute/vol solution

•         Solution = solute + solvent

•         Solution= 1000 ml + 2000 ml = 3000 ml

•         1000 ml /3000 ml= 33.3 %

11. If 1000 ml of H3PO4 (5 moles) is added to 2000ml of water, calculate Molarity

•         Solution=1000 + 2000=3000ml

•         3000ml/1000 ml/L=3 L

•         Molarity=mol solute/L solution

•         5 mole/3L= 1.667 M

12. If 1000 ml of H3PO4 (5 moles) isadded to 2000ml of water, calculate Normality.

•         Normality= equiv/Liters solution

•         H+1 = 3 for every mole

•         3  (5 moles) = 15 eq

•         3000ml/1000 ml/1L=3 L

•         15 eq/3 L =  5 N

13. If 1500 ml of H2SO4 (3mole) is added to 2500ml of water, calculate % volume, Molarity, and Normality.

•         % volume=vol. solute/Vol. solution

•         Solution=1500 + 2500=4000ml

•         1500/4000=37.5 %

•         Molarity=mol solute/L solution

•         1500 ml + 2500 ml= 4000ml

•         4,000ml/1000= 4 L

•         3mole/4L= 0.75 M

•         Normality= equiv/Liters solution

•         H2= 2 hydronium ions

•         2(3moles) =6 eq

•         1500+2500= 4000 ml

•         4,000ml/1000= 4 L

•         6 eq/4 L= 1.5 N

14. What is the new concentration of a 200 ml solution, if at 50 ml it was 8 N?

•         Conci(volumei) =Concf(volumef)

•          8 N(50 ml) = Concf(200 ml)

•         Solve for Concf

•          8 N(50 ml) /(200 ml) = Concf(200 ml)/(200 ml)

•         Concf=2 N

15. What is the new concentration if a 200 ml solution of 60 % had an addition 100 ml added to it?

•         Conci(volumei) =Concf(volumef)

•          60 %(200 ml) = Concf(300 ml)

•         Solve for Concf

•         60 %(200 ml) /(300 ml)=Cncf(300 ml)/(300 ml)

•         Concf= 40 %

16.  What is the new concentration of a 40 L solution, if at 10 L it was 5 m?

•         Conci(volumei) =Concf(volumef)

•          5 m(10 L) = Concf(40 L)

•         Solve for Concf

•         5 m(10 L) /(40 L) =Concf(50 L)/(40 L)

•         Concf= 1.25 m

17. What is the new concentration if a 4 L solution of 2 M had an addition 2 L added to it?

•         Conci(volumei) =Concf(volumef)

•          2 M(4 L) = Concf(6 L)

•         Solve for Concf

•          2 M(4 L) /(6 L) = Concf(6 L)/(6 L)

•         Concf= 1.33 M

 18. If 500 ml of H2SO4 (1mole) is added to 1000ml of water, calculate: % volume, Molarity, and Normality

•         % volume=vol. solute/Vol. solution

•         Solution =Solute + solvent

•         Solution=500 + 1000=1500ml

•         500/1500(100)=33.3%

•         Molarity=mol solute/L solution

•         Solution =Solute + solvent

•         Solution=500 + 1000=1500ml

•         1500ml/1000ml=1.5L

•         1mole/1.5 L= 0.67M

•         Normality= equiv/Liters solution

•         # of moles(#of H)

•         1 mole(2 H)=2 equiv.

•         1000+500=1500

•         1500ml/1000=1.5 L

•         equiv/Liters solution =2/1.5=1.33 N

19. What is the molality and % mass if 1000 ml of alcohol(.9 g/ml density) has 1 mole of Magnesium Chloride dissolved in it.

•         Molality=moles solute/Kg solvent

•         1 mole

•         1000 ml(.9 g/ml)=900 g

•         900 g/1000g/kg=.9 kg

•          1mole/.9 Kg=1.1 m

•         % mass=mass solute/Mass of solution

•         1000 ml(.9 g/ml)=900 g

•         MgCl2  =24 + 71= 95 g

•         900 + 95 g =995 g

•         Solute/solution=95/995 =

•         9.55 %

20. If 500 ml of HCl (3 moles) is added to 2000 ml of water, calculate  % volume.

•         % volume=vol.

•         solute/vol solution

•         solution=solute+solvent=500+2000= 2500ml

•         500/2500= 20 %

•         Molarity=mol solute/Liters solution

•         Solution=500 + 2000=2500ml

•         2500ml/1000ml/L=2.5 L

•         3 mole/2.5 L= 1.2 M

•         Normality= equiv/Liters solution

•         H= 1 equiv.

•         1 equiv (3 moles)=1X3=3eq

•         2500ml/1000=2.5 L

•         3/2.5=  1.2 N

21. What is the molality if 2000 ml of water has 240 g of Beryllium  Chloride dissolved in it.

•         Molality=moles solute/Kg solvent

•         Be= 9  Cl= 35.5  BeCl2 =9+71=80 g/mol

•         240g/80g/mol= 3 moles

•         2000 ml(1g/ml)=2000 g

•         2000 g/1000 g/1 kg= 2 kg

•          3 moles/2 Kg=1.5 m

22. What is the  % mass if 2000 ml of water has 240 g of Beryllium  Chloride dissolved in it.

•         % mass=mass solute/Mass solution

•         2000 ml(1g/ml)=2000 g

•         2000 g +240 g = 2240 g

•         240 g/2240 g(100 %)

•         10.7 %

23. What is the molality if 2000 ml of camphor has 400 g of Sodium Phosphide dissolved in it?

•         Molality=moles solute/Kg solvent

•         Na= 23  P= 31  Na3P =69+31=100 g/mol

•         400g/100g/mol= 4 moles

•         2000 ml(.6 g/ml)=1200 g

•         1200 g/1000 g/1 kg= 1.2 kg

•          4 moles/1.2 Kg= 3.3 m

What is the new freezing and boiling points?

•         Freezing point- m(Kf/m)

•         10 0C –3.3 m(37.7 0C /m)

•         10 0C –124.41 0C = -114.41 0C

•         Boiling point+ m(Kb/m)

•         48 0C +3.3 m(5.95 0C /m)

•         48 0C +19.635 0C = 67.635 0C

24. What is the molality if 3000 ml of  Benzene has 564 g of Cesium Oxide dissolved in it?

•         Molality= moles solute/Kg solvent

•         Cs=133  O=16  Cs2O = 266+16=282 g/mol

•         564g/282g/mol= 2 moles

•         3000 ml(.81 g/ml)= 2430 g

•         2430 g/1000 g/1 kg= 2.43 kg

•          2 moles/2.43 Kg= 0.823 m

What is the new freezing and boiling points?

•         Freezing point- m(Kf/m)

•         -72 0C –.823m(5.12 0C /m)

•         -72 0C –4.214 0C = -76.214 0C

•         Boiling point+ m(Kb/m)

•         81 0C + .823 m(2.53 0C /m)

•         81 0C +2.08 0C = 83.08 0C