Question

Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some unknown mass of sucrose is dissolved in 150g of water and this solution has a freezing/melting point of -0.56°C, calculate the mass of sucrose dissolved. Kfp for water is 1.86°C/m. [must show work including units to receive credit].

Answer #1

Given that Kfp for water is 1.86°C/m

depression in freezing point = normal freezing/melting point 0.0°C - freezing/melting point of solution

= 0.0 °C - (- 0.56°C )

= 0.56 °C

depression in freezing point = i*m*Kf

for nonionic solute ; i = 1

First calculate the molality of sucrose as follows:

0.56 °C= 1 * m * 1.86°C/m

m = 0.30 mole / kg

mass of water = 150 g or 0.150 kg

now calculate the number of moles of sucrose as follows:

molality = number of moles / solvent inKG

number of moles = 0.30 mole / kg *0.150 kg

= 0.045 mole sucrose

Molar mass of Sucrose is **342.2965** g/mol.

amount of sucrose in g = number of mole * molar mass

= 0.045 mole sucrose ***342.2965** g/mol

= 15.40 g sucrose

The normal freezing point of water,
H2O is 0.00 °C and its
Kfp value is 1.86 °C/m.
Assuming complete dissociation of the electrolyte, if
10.87 grams of cobalt(II) nitrate
(Co(NO3)2,
183.0 g/mol) are dissolved in
289.5 grams of water what is the
freezing point of the solution?

1. Estimate the freezing point of 150 cm3 of water
sweetened with 7.5g of sucrose
(C12H22O11). (Kf for water is 1.86
K kg mol−1 ; density of water is 1g cm−3
)
Show work

Calculate the freezing point and boiling point of each of the
following solutions: the freezing point of the solution: 174 g of
sucrose, C12H22O11, a
nonelectrolyte, dissolved in 1.35 kg of water
(Kf=1.86∘C) Express your answer using one decimal
place.

1) What will the solution vapor pressure be for a solution of
11.07 mol of sucrose (C12H22O11)
dissolved in 1.000L of water at 20.0 °C? The vapor pressure of pure
water at 20.0 °C is 17.5 torr. (Show work)
2) Calculate the freezing point of a solution containing 0.43 of
sucrose in 150 mL of water. Kf = 1.86 °C/m (Show
Work)

The normal freezing point of water is 0.0 oC. At this temperature
the density of liquid water is 1.000 g/ml and the density of ice is
0.917 g /ml. The increase in enthalpy for the melting of ice at
this temperature is 6010 J/mol. What is the freezing point of water
at 200 atmospheres?

he normal freezing point of water is 0.0 degrees celsius. at
this temperature the density of liquid water is 1.000 g/mL and
density of ice is 0.917 g/mL. the increase in enthalpy for the
melting ice at this temperature is 6010 J/mol. What is the freezing
point of the water at 200 atms?

Since pure water boils at 100.00 ∘C, and since the addition of
solute increases boiling point, the boiling point of an aqueous
solution, Tb, will be Tb=(100.00+ΔTb)∘C Since pure water freezes at
0.00 ∘C, and since the addition of solute decreases freezing point,
the freezing point of an aqueous solution, Tf, will be
Tf=(0.00−ΔTf)∘C
What is the boiling point of a solution made using 735 g of
sucrose, C12H22O11, in 0.225 kg of water, H2O?
What is the freezing point...

The freezing point of water H2O is 0.00°C at 1 atmosphere. A
nonvolatile, nonelectrolyte that dissolves in water is antifreeze
(ethylene glycol).
How many grams of antifreeze, CH2OHCH2OH (62.10 g/mol), must be
dissolved in 297.0 grams of water to reduce the freezing point by
0.400°C ?
_____ g antifreeze.

A certain solution of benzoic acid in benzene has a freezing
point of 3.1°C and a normal boiling point of 82.6°C. Explain these
observations and suggest structures of the solute molecules at
these two temperatures. Kfp = 5.12°C/m and
Kbp = 2.53°C/m.

A certain substance X has a normal freezing point of -6.4 C and
a molal freezing point depression constant Kf= 3.96
degrees C.kg.mol-1. A solution is
prepared by dissolving some urea ((NH2)2CO)
in 950. g of X. This solution freezes at -13.6 C. calculate the
mass of urea that was dissolved. Round your answer to 2 significant
digits.

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