Question

What is the percent of salt in a salt solution that has freezing
point of -6.7 C ? Freezing point of water 0 C

Answer #1

Data given: Depression in freezing point = Difference between the freezing point of water and that of salt solution = 6.7 degree celsius

And cryoscopic constant for water = 1.86

The depression in freezing point is related to molality of the solution by the equation

depression in freezing point = K_{f} molality therefore,
6.7 = 1.86 x molality

Molality = 6.7/1.86 = 3.60 mol kg^{-1} of the
solvent

This means in the 1 kg of water contains 3.6 mol of salt = 3.6
x58.8 g of salt(58.5 g mol^{-1} is the molar mass of
salt)

Thus 1 kg of water contains 210.7 g of salt

The percent of salt in the solution is therefore, 210.7x100/1000= 21.07%

what is the freezing point of an aqueous 200m (NH4)PO4 salt
solution? The Kf of water 1.86 C/m. Assume complete dissociation of
the soluble salt.

What is the normal boiling point of an aqueous solution that has
a freezing point of 1.04 oC. Kf for water 1.86 oC/m (oC-kg/mol).
Hint: Calculate the molality from the freezing point depression and
use it to calculate the normal boiling point.

What is the freezing point of an aqueous solution that boils at
107.4 ∘C? You may assume that the boiling and freezing points of
water are 100 ∘C and 0 ∘C, respectively

A solution of 5.00 g of sodium chloride in 1.00 kg of water has
a freezing point of –0.299°C. What is the actual (experimental) van
’t Hoff factor for this salt at this concentration?
Kf(water) =
1.86°C/m

To use freezing-point depression or boiling-point elevation to
determine the molal concentration of a solution.
The freezing point, Tf, of a solution is lower than the
freezing point of the pure solvent. The difference in freezing
point is called the freezing-point depression, ΔTf:
ΔTf=Tf(solvent)−Tf(solution)
The boiling point, Tb, of a solution is higher than the
boiling point of the pure solvent. The difference in boiling point
is called the boiling-point elevation, ΔTb:
ΔTb=Tb(solution)−Tb(solvent)
The molal concentration of the solution, m,...

Calculate the boiling point of a solution of NaCl that has a
freezing point of -0.3720 °C. Assume complete dissociation. Kf
water = 1.86 °C/m Kb water = 0.512 °C/m
A. 100.1 °C
B. 99.1 °C
C. 101.1 °C
D. 98.9 °C
E. 105 °C

A sample of acetic acid solution has a freezing point depression
of 2.0 C. What is the e estimated percent by mass of acetic acid in
the acetic acid solution? Molar mass of acetic acid
C2H3O2=60g/mol.

The experimentally measured freezing point of a 1.05 m aqueous
solution of AlCl3 is -6.25°C. The freezing point depression
constant for water is Kf = 1.86°C/m. Assume the freezing point of
pure water is 0.00°C.
Part 1 What is the value of the van't Hoff factor for this
solution?
Part 2 What is the predicted freezing point if there were no ion
clustering in the solution?

What is the freezing point of a solution prepared by adding 50.0
g of NaCl to 250. g of pure water? (Water has a kfp =1.86
oC/m)

Each question is and was separated by an asterisk.
*The freezing point of water is
0.00°C at 1 atmosphere.
If 14.18 grams of cobalt(II)
bromide, (218.7 g/mol), are dissolved in
271.6 grams of water ...
The molality of the solution is m.
The freezing point of the solution is °C.
*The freezing point of water
H2O is 0.00°C at 1
atmosphere.
How many grams of sodium nitrate
(85.00 g/mol), must be dissolved in
249.0 grams of water to reduce
the freezing point...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 1 minute ago

asked 5 minutes ago

asked 20 minutes ago

asked 30 minutes ago

asked 35 minutes ago

asked 50 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago