A solution is made of 2.00g sucrose in 100.00g H2O (molecular weight of sucrose is 342g/mol). Calculate the a.) boiling point and b.) freezing point of the solu

Question

I really need help...

Answer 1

Hello Maria

 

This question is based on the colligative properties of a solution where the solute will cause an increase in the boiling point and a decrease in the freezing point of the pure solvent.  For these calculations, we need to use a concentration unit called molality, which is a ratio of the moles of solute per kg of solvent.

 

In the above question, sucrose is your solute while water is the solvent.

 

First we need to calculate the # of moles of sucrose using its molar mass.

 

2.00 g sucrose x 1 mol sucrose / 342 g sucrose  =  0.00585 mol sucrose

 

Now we can calculate the molality (m) of the solution.

 

m = moles of solute / kg of solvent

 

Since water is the solvent, and its mass is 100.00 g, use the conversion factor, 1 kg = 1000 g to obtain the mass of water in kg, which = 0.10000 kg (to the correct # of significant figures)

 

m  =  0.00585 mol sucrose / 0.100000 kg H₂O  =  0.0585 m

 

To calculate the boiling point of the solution, we need to use the equation,

 

ΔT_b  =  K_b  x  m

 

K_b is the boiling point constant for H₂O, and its value is 0.512 ºC / m.

 

ΔT_b = 0.512 ºC / m x 0.0585 m  =  0.0300 ºC

 

Since ΔT_b = T_b - T_bº, where T_b is the boiling point of the solution and T_bº is the boiling point of the solvent (H₂O), which = 100.00 ºC, rearrange to solve for T_b.

 

T_b = ΔT_b + T_bº  = 0.0300 + 100.00 = 100.03 ºC

 

Similarly, to calculate the freezing point of the solution, you would use the equation:

 

ΔT_f = K_f x m

K_f is the freezing point constant for H₂O, and its value is 1.86 ºC / m.

 

ΔT_f = K_f x m  =  1.86 ºC / m x 0.0585 m = 0.109 ºC

 

Since ΔT_f = T_fº - T_f, where T_f is the freezing point of the solution and T_fº is the freezing point of H₂O, which = 0.00 ºC, rearrange to solve for T_f.

T_f = T_fº - ΔT_f  = 0.00 - 0.109 = - 0.109 ºC

Answer 2

Adding sucrose to the water will increase water's boiling point (to greater than 100 C) and decrease its freezing point (to less than 0 C). 

 

Boiling Point elevation is given by:

 dT = kb * m  

where dT is the change in temperature, kb is the boiling point elevation constant for water and m is the molality of the solution.

 

for water, kb = 0.512 C/m

 

molality of this solution is:

 

moles sucrose / kg water 

 

moles sucrose = 2 g sucrose / (342 g/mol)

kg water = 100g/(1000g/kg)

 

molality = 0.05847, round to 0.06 mol/kg

 

now, dT = 0.512 C/m * 0.06 mol/kg = 0.03 C

 

so the boiling point of the solution is 100C + 0.03 C = 100.03 C. 

 

 

Calculating the freezing point is the same procedure except you use Kf which for water is 1.86. You will also subtract the dT from 0C since the sucrose lowers the freezing point of the water. 

 

Hope that helps! Let me know if anything is unclear. 

 

 

Answer 3

The equations needed are ΔT_f= -K_f*m_solute and ΔT_b= K_b*molality.

 

mol_solute= 2.00 g/ 342 g/mol= 0.00584 mole

Molality= 0.0584 

 

K_f= 1.86

K_b= 0.512

 

For boiling point elevation:

 

ΔT_b= 0.512*0.0584

ΔT_b= 0.0299 

 

Now we use the that in this equation: T_solution= T_puresolvent+ΔT_b

T_solution= 100+0.0299= 100.03 °C

 

For the freezing point, it's pretty the same thingjust using the K_f value.

 

ΔT_f=-1.86*0.0584

ΔT_f= - 0.109

 

T_solution= 0-0.109= -0.109ºC