Question

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?

Answer #1

**Solution.**

The curve along which a pure crystal and its melt maintain equal chemical potentials is given by an equation:

Let's suppose we have 1 mole of water, which weighs 18 g.

The volume of a liquid is V_{l} = 18/1 = 18 mL;

the volume of a solid is Vs = 18/0.917 = 19.63 mL;

the difference is 18-19.63 = -1.63 mL/mol.

T = 271.66 K, or -1.49 °C.

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?

Assume that the density of solid water is 920 kg m-3 and that of
liquid water is 1000 kg m-3. Calculate the change of the melting
temperature, T, of ice for the process of changing pressure from
100 kPa to 1850 kPa. Use fusH = 6010 J mol-1 for water.

The normal melting point of H2O is 273.15 K, and Hfusion= 6010
J mol–1. Calculate the decrease in the normal freezing point at 100
and 500 bar assuming that the density of the liquid and solid
phases remains constant at 997 and 917 kg m–3, respectively.

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].

7.1 Assume that the density of solid water is 920 kg m-3 and
that of liquid water is 1000 kg m-3. Calculate the change of the
melting temperature, T, of ice for the process of changing
pressure from 100 kPa to 1850 kPa. Use fusH = 6010 J mol-1 for
water.
7.2 At two temperatures, 85 K and 135 K, the vapor pressures of
a certain liquid were found to be 540 Torr and 830 Torr,
respectively. From this information,...

The freezing point depression of a solution of nitrobenzene and
a nonionic unknown was used to determine the molar mass of the
unknown. Time-temperature data for the cooling of nitrobenzene and
for the cooling of a solution containing 50.0 g of nitrobenzene and
5.00 mL of a nonionic liquid unknown. Density of the unknown is
0.714 g mL^-1. And the Kf of nitrobenzene is 6.87 degrees C Kg
mol^-1.
What is the freezing point of the unknown solution? I
know...

35 mL of water is at 87 degrees Celsius. This water is heated,
becomes steam and then continues heating to a temperature of 106
degrees Celsius. How much heat was absorbed due to this
process?
Cwater= 4.186 J/g
C Csteam= 2.02 J/g
C Vaporization:
40.65 kJ/mol

What is the difference in temperature between the freezing point
of water and the normal boiling point of water on the Fahrenheit,
Celsius scale and on the Kelvin scale? Explain how you figured this
out.

1. At the normal melting point of ice
∆Hfus= 6.007 kJ mol−1 and
∆Sfus= 22.00 J K−1
mol−1.
a) What is ∆Gfusat the normal melting
point? Look at your answer. Is it correct? Why or why not?
b) Determine ∆G for freezing water at 1 atm and
-10oC assuming that
∆Hfusand
∆Sfusdo not change much over the 0 →
−10oC temperature range.
c) Determine ∆G for freezing water at 1 atm and
-10oC assuming that
∆Hfusdoes not change much over...

The normal melting point of substance A is 320C. Melting
heat is a function of temperature. Determine the melting
temperature of substance A at 10 bar.
Data :
Density of liquid A = 10200
kg/m3
Density of Solid A = 12000 kg/m3
For liquid A CpL =
30,0-3,1 x 10-3T (J/mol.K)
For solid A CpS = 20,0 +
9x10-3T (J/mol.K)
Tref =
320oC
ΔHreferime = 4900
J/mol
MA = 250 kg/kmol

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 13 minutes ago

asked 13 minutes ago

asked 17 minutes ago

asked 19 minutes ago

asked 24 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago