Question

a. An aqueous solution is made by dissolving 18.4 grams of iron(III) bromide in 314 grams of water.  

The molality of iron(III) bromide in the solution is ________m.

b.In the laboratory you are asked to make a 0.473 m chromium(II) acetate solution using 425 grams of water.  

How many grams of chromium(II) acetate should you add?  
_________ grams.

c. The boiling point of diethyl ether CH₃CH₂OCH₂CH₃ is 34.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is chlorophyll.

If 10.38 grams of chlorophyll, C₅₅H₇₂MgN₄O₅ (893.5 g/mol), are dissolved in 236.2 grams of diethyl ether.

The molality of the solution is ______ m.

The boiling point of the solution is______ °C.

d. The freezing point of benzene C₆H₆ is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT .  

If 14.38 grams of DDT, C₁₄H₉Cl₅ (354.5 g/mol), are dissolved in 276.5 grams of benzene ...

The molality of the solution is _____ m.

The freezing point of the solution is _________°C.

Answer 1

a. An aqueous solution is made by dissolving 18.4 grams of iron(III) bromide in 314 grams of water.  

Number of moles of iron(III) bromide = 18.4 g/ 295.56 g/mol

= 0.062 mol iron(III) bromide

molality of iron(III) bromide = number of moles / solvent in Kg

= 0.062 mol iron(III) bromide /0.314 kg

=0.197 m
The molality of iron(III) bromide in the solution is 0.197 m .

b.In the laboratory you are asked to make a 0.473 m chromium(II) acetate solution using 425 grams of water.  

Given that m = 0.473 m = number of moles / solvent in Kg

Number of moles = 0.473 moles / kg * 0.425 Kg

=0.20 mol

molar mass of chromium(II) acetate= 229.13 g/mol

Now calculate the amount of 0.20 mol of chromium(II) acetate :

229.13 g/mol*0.20 mol

= 46.1 g
How many grams of chromium(II) acetate should you add?  
46.1 grams.

c. The boiling point of diethyl ether CH₃CH₂OCH₂CH₃ is 34.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is chlorophyll.

If 10.38 grams of chlorophyll, C₅₅H₇₂MgN₄O₅ (893.5 g/mol), are dissolved in 236.2 grams of diethyl ether.

Number of moles = 10.38 grams of chlorophyll/ 893.5 g/mol= 0.012 mol

Mass of solvent in kg= 0.2362 kg

Molality = 0.012 mol/ 0.2362 kg=0.051 m
The molality of the solution is _0.051 _____ m.

Given that

The boiling point of diethyl ether CH₃CH₂OCH₂CH₃ is 34.50°C
The value of Kb for diethyl ether is 1.824 °C kg/mol

Change in BP = Kb m where Kb is the boiling point elevation constant for diethyl ether

Change in BP = 1.824 °C kg/mol*0.051 mol/kg

Change in BP = 0.093 °C

Boiling point of solution = 34.50°C+ 0.093 °C=34.593 °C
The boiling point of the solution is34.593 °C

d. The freezing point of benzene C₆H₆ is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT .  

If 14.38 grams of DDT, C₁₄H₉Cl₅ (354.5 g/mol), are dissolved in 276.5 grams of benzene ...

Number of moles = 14.38 grams of DDT / 354.5 g/mol= 0.041 mol

Mass of solvent in kg= 0.2765 kg

Molality = 0.041 mol/ 0.2765 kg=0.15 m
The molality of the solution is _0.15_____ m.

Given that

The freezing point of benzene C₆H₆ is 5.50°C

The value of Kf for benzene is 4.90 C/m

dT = kf x m = 4.90 C/m x 0.15 m = 0.735 C

The f.p. is 5.50 - 0.735 = 4.765 C

The freezing point of the solution is _4.765 _°C.