Question

he normal freezing point of water is 0.0 degrees celsius. at this temperature the density of...

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?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

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 Vl = 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.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The normal freezing point of water is 0.0 oC. At this temperature the density of liquid...
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...
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...
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....
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...
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...
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...
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...
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...
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....
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
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT