When 1.6 mol of KCl is dissolved in 1 kg of water (K = 2 K/molal), the freezing point of the resulting solution will be:

You use the formula for freezing point depression.

Explanation:

EXAMPLE

What is the freezing point depression caused by adding 31.65 g of sodium chloride to 220.0 g of water.  

K

f

for water is  

1.86 °C⋅kg⋅mol

-1

.

Solution

The formula for freezing point depression expression is

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

Δ

T

f

=

i

K

f

m

a

a

∣

∣

−−−−−−−−−−−−−−−  

where

Δ

T

f

is the freezing point depression,

i

is the van’t Hoff factor,

K

f

is the molal freezing point depression constant for the solvent, and

m

is the molality of the solution.

Step 1: Calculate the molality of the NaCl

moles of NaCl

=

31.65

g NaCl

×

1

mol NaCl

58.44

g NaCl

=

0.5416 mol NaCl

mass of water

=

220.0

g H

2

O

×

1

kg H

2

O

1000

g H

2

O

=

0.220 kg H

2

O

m

=

moles of NaCl

kilograms of water

=

0.5416 mol

0.220 kg

=

2.46 mol/kg

Step 2: Determine the van't Hoff factor

The van't Hoff factor,  

i

, is the number of moles of particles obtained when 1 mol of a solute dissolves.

Nonelectrolytes such as sugar do not dissociate in water. One mole of solid sugar gives one mole of dissolved sugar molecules.

For nonelectrolytes,  

i

=

1

.

Electrolytes such as  

NaCl

completely dissociate into ions.

NaCl

→

Na

+

+

Cl

-

One mole of solid  

NaCl

gives two moles of dissolved particles: 1 mol of  

Na

+

ions and 1 mol of  

Cl

-

ions. Thus, for  

NaCl

,

i

=

2

.

Step 3: Calculate  

Δ

T

f

Δ

T

f

=

i

K

f

m

=

2

×

1.86 °C⋅

kg⋅mol

-1

×

2.46

mol⋅kg

-1

=

9.16 °C

The freezing point depression is 9.16 °C.

Here's a video on how to calculate freezing point depression.You use the formula for freezing point depression.

Explanation:

EXAMPLE

What is the freezing point depression caused by adding 31.65 g of sodium chloride to 220.0 g of water.  

K

f

for water is  

1.86 °C⋅kg⋅mol

-1

.

Solution

The formula for freezing point depression expression is

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

Δ

T

f

=

i

K

f

m

a

a

∣

∣

−−−−−−−−−−−−−−−  

where

Δ

T

f

is the freezing point depression,

i

is the van’t Hoff factor,

K

f

is the molal freezing point depression constant for the solvent, and

m

is the molality of the solution.

Step 1: Calculate the molality of the NaCl

moles of NaCl

=

31.65

g NaCl

×

1

mol NaCl

58.44

g NaCl

=

0.5416 mol NaCl

mass of water

=

220.0

g H

2

O

×

1

kg H

2

O

1000

g H

2

O

=

0.220 kg H

2

O

m

=

moles of NaCl

kilograms of water

=

0.5416 mol

0.220 kg

=

2.46 mol/kg

Step 2: Determine the van't Hoff factor

The van't Hoff factor,  

i

, is the number of moles of particles obtained when 1 mol of a solute dissolves.

Nonelectrolytes such as sugar do not dissociate in water. One mole of solid sugar gives one mole of dissolved sugar molecules.

For nonelectrolytes,  

i

=

1

.

Electrolytes such as  

NaCl

completely dissociate into ions.

NaCl

→

Na

+

+

Cl

-

One mole of solid  

NaCl

gives two moles of dissolved particles: 1 mol of  

Na

+

ions and 1 mol of  

Cl

-

ions. Thus, for  

NaCl

,

i

=

2

.

Step 3: Calculate  

Δ

T

f

Δ

T

f

=

i

K

f

m

=

2

×

1.86 °C⋅

kg⋅mol

-1

×

2.46

mol⋅kg

-1

=

9.16 °C

The freezing point depression is 9.16 °C.

Here's a video on how to calculate freezing point depression.h

i believe the answer is true.

here are some of the ways in which chemical and nuclear reactions differ.

explanation: