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The addition of a nonvolatile solute to a solvent causes the boiling point of the solvent to increase and the freezing point of the solvent to decrease. 

Regents Questions-Follow the Regents link to check the answer

Jan 2003-19 What occurs when NaCl(s) is added to water

(1) The boiling point of the solution increases, and the freezing point of the solution decreases.
(2) The boiling point of the solution increases, and the freezing point of the solution increases.
(3) The boiling point of the solution decreases, and the freezing point of the solution decreases.
(4) The boiling point of the solution decreases, and the freezing point of the solution increases.

June 2003-23 At standard pressure when NaCl is added to water, the solution will have a

(1) higher freezing point and a lower boiling point than water
(2) higher freezing point and a higher boiling point than water
(3) lower freezing point and a higher boiling point than water
(4) lower freezing point and a lower boiling point than water

Jan 2007-19 Compared to a 2.0 M aqueous solution of NaCl at 1 atmosphere, a 3.0 M aqueous solution of NaCl at 1 atmosphere has a
(1) lower boiling point and a higher freezing point
(2) lower boiling point and a lower freezing point
(3) higher boiling point and a higher freezing point
(4) higher boiling point and a lower freezing point

Jun 2009-20 Compared to the freezing point and boiling point of water at 1 atmosphere, a solution of a salt and water at 1 atmosphere has a

(1) lower freezing point and a lower boiling point
(2) lower freezing point and a higher boiling point
(3) higher freezing point and a lower boiling point
(4) higher freezing point and a higher boiling point

Advanced Chemistry Calculations

Table 1

Example 1- van't Hoff   i= 1 (nonelectrolytes)

A solution is made by dissolving 10.20 grams of glucose (C₆H₁₂O₆) in 355 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

 DT_f=imK_f

i=1 (see table 2);    m= mol/kg

moles=10.20g (1mol/180g)=0.0556mol                kg=355g(1 kg/1000g)=0.355kg

m= mol/kg= 0.0556 mol/0.355kg= 0.156mol/kg;      

K_f=-1.86^oC/m (see table 1)

DT_f=1(0.156mol/kg)(-1.86^oC/m)= -0.291^oC

Table 2-Van't Hoff Factors (i)

i=van't Hoff= the number of particles a formula breaks up into C6H12O6(s) ==>C6H12O6(aq)    (1 mole of particles)       C6H12O6 =1 NaCl(s) ==> Na+(aq) + Cl-(aq)    (2 moles of particles)      NaCl=2 CaCl2(s) ==> Ca2+(aq) + 2Cl-(aq)   (3 moles of particles)    CaCl2=3 More on the Van't Hoff Factor

Table 3

For every mole of particles added to1 kg of pure water the freezing point decreases by 1.86oC and the boiling point increases by 0.52oC

Example 2- van't Hoff   i= 3 (electrolytes)

Determine the freezing point of a solution made from 3.0 kgs of water and 2.5mol of CaCl₂.

 DT_f=imK_f

i=3 (see table 2);     m= mol/kg= 2.5 mol/3.0 kg= 0.83mol/kg;       K_f=-1.86^oC/m (see table 1)

DT_f=3(0.83mol/kg)(-1.86^oC/m)= -4.63

Freezing point= 0.00^oC -4.63^oC=-4.63^oC

Example 3- Boiling Point elevation; van't Hoff  i=2 (electrolyte)

Determine the boiling point of a solution made from 2,000.g of water and 58.5g mol of NaCl.

 DT_b=imK_b

i=2 (see table 2);  

m= mol/kg

mol NaCL= 58.5g(1mol/58.5g)=1.00mol NaCl    kg=2,000.g(1 kg/1000g)=2.000kg

m= mol/kg= 1.00 mol/2.000kg= 0.5mol/kg;       K_b=+0.52^oC/m (see table 1)

DT_b= (2)(0.5mol/kg)(+0.52^oC/m) =+0.52^oC

Boiling point= 100.^oC +0.52^oC=100.52^oC

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