Question

After adding a solute freezing point of solution decreases to -0.186 calculate delta Tb if Kf =1.86 and Kb =0.521

Answer 1

Tb = Kbm Kb H2O = 0.52 degrees C/m  

Tf = Kfm Kf H2O = 1.86 degrees C/m  

show me the work  

1. What is the new boiling point if 25 g of NaCl is dissolved in 1.0 kg of water?  

2. What is the freezing point of the solution above?  

3. What are the new freezing and boiling points of water if 50. g of ethylene glycol (molar mass = 62 g/mol) is added to 50. g of water?  

4. When 4.0 g of a nonelectrolyte is added to 25 g of water, the new freezing point is – 2.5 oC. What is the molar mass of the unknown compound?

Answer 2

Recall that in colligative properties, solution concentration is measured in MOLALITY (with emphasis to MOLAL) not molarity. So the formula for molality is 

molality = mol solute / kg solvent 

In the first problem, we are given the amount of solute in grams. Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. 

25 g NaCl / 58.5 g/mol = 0.427 mol 

Then, use the formula for molality. 

molality = mol solute / kg solvent 
= 0.427 / 1 
= 0.427 m 

Use now the formula (we're asked for the boiling point here!) 

delta Tb = Kbm 
= (0.52)(0.427) 
= 0.22C 

Recall that water boils at 100C. When a nonvolatile solute is added to water, the boiling point of water increases. We now add 0.22 C to 100 C so we know the new boiling point of the solution. 

100 C + 0.22 C = 100.22 C (new boiling point) 

For the second question, we use the formula (this time, the freezing point!)....let's use the molality from the first problem as our molality. 

delta Tf = Kfm 
= (0.427)(1.86) 
= 0.79C 

Recall that water freezes at 0 C. We then subtract 0.79 from 0 C. Remember that adding a nonvolatile solute to water decreases its freezing point. 

0 - 0.79 = -0.79 C 

In the third question, we also get the molality first like what we did in the previous problem (the first and second problems are linked). 

Converting 50 g ethylene glycol by dividing by the molar mass gives 

50 g / 62 g/mol = 0.806 mol 

We convert first 50 g H2O (that's our solvent) to kg. Recall that amount of solvent is given in kg. Dividing by 1000, we get 0.05 kg. 

Get the molality. 

m = 0.806 mol / 0.05 kg 
= 16.12 m 

Then, use the formula, delta Tb = Kbm to calculate for the boiling point of the new solution. 

delta Tb = Kbm 
= (0.52)(16.12) 
= 8.38C 

Add this to the boiling point of water, 100 C. 

100 C + 8.38 C = 100.38 C (new boiling point) 

We use the same molality in calculating for the freezing point. 

delta Tf = Kfm 
= (1.86)(16.12) 
= 30.0 C 

Subtracting 30.0 C from 0 C gives us -0.30 C. 

The fourth problem needs a bit thorough discussion. We need first to know delta Tf (actually, we know already the freezing point) but we need to know by how much the freezing point of the solution decreased. We subtract -2.5 C from 0 so we'll know the value of delta Tf. 

delta Tf = 2.5 C 

Now, we derive the formula delta Tf = Kfm for the molality. We need to know the molality. I'll explain why later on. 

m = delta Tf / Kf 
= 2.5 / 1.86 
= 1.34m 

Recall that amount of solvent is given in kg. We convert 25 g H2O to kg. Dividing by 1000, we get 0.025 kg. 

We now multiply 0.025 kg by the molality so we'll know how many moles there are in the solution. We need this in calculating the molar mass of the compound. 

0.025 kg * 1.34 mol / kg = 0.0335 mol 

To get the molar mass of the compound, we divide 4.0 g by 0.0335 mol. 

molar mass = grams of compound / moles of compound 
= 4.0 g / 0.0335 mol 
= 119 g/mol