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# Tempurature Dependence of Salt Solubility

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I don’t understand how to make this graph...what is it supposed to look like? I’ve tried and just can’t make my computer work it...I also don’t understand what my experimental mass and official masses are supposed to be... please help!
Data: List all measurements and include proper labels.
Test tube Crystallizing Temperature (degrees C)
Solute (g/10ml) KCl NH4Cl
3.5 23.94 degrees C -----
4.0 39.69 degrees C 23.12 degrees C
4.5 57.06 degrees C 34.87 degrees C
5.0 72.22 degrees C 47.55 degrees C
6.0 --- 68.53 degrees C
Calculations/Interpretations: Show all math performed (give the formula, show your setup, and give the result. Use proper significant figures and include proper labels. Also, answer ALL questions asked in the lab.
Test tube
Solute (g/10ml) Solute (g/ 100ml)
3.5
4.0
4.5
5.0
6.0
Official points to plot.
Temperature (C) KCl (g/100ml) NH4Cl (g/100ml)
30 37.1 41.6
50 42.9 50.4
70 48.5 59.9
90 53.8 70.4
Plot these 4 sets of data points on a graph:
1. Experimental KCl solubility
2. Experimental NH4Cl solubility
3. Official KCl solubility
4. Official NH4Cl solubility
Plot temperature (deg C) on the Y-axis (vertical) and solubility (g/100ml) across the X-axis (horizontal).
The temperature range should be from 0 to 100 deg C. The solubility range should be from 0g/100ml to 100 g/100ml.
From Graphs:
KCl NH4Cl
Experimental mass at 60 C:
Official mass at 60 C:
% difference:
(a) Why is it important to not allow the test solutions to boil?
It would be a poor variable to cause a change the volume of the solution, boiling would alter this far more than evaporation from an 80 degree solution.
(b) Why is it better to determine the initial crystallization temperature during cool down rather than the temperature at which all of the salt dissolves?
It is harder to see the point at which all the salt has dissolved and it dissolved through agitation as well as increase in temperature, i.e. as you would warm it up it doesn't all dissolve for the saturation point of that temperature immediately. It is far easier to note the crystallization point when cooling.

https://brainmass.com/chemistry/stoichiometry/tempurature-dependence-salt-solubility-282085

#### Solution Summary

This solution explains how to solve severla problems dealing with temperature dependence of salt solubility.

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## General Chemistry

A saturated solution _______________.

a. cannot be attained
b. contains dissolved solute in equilibrium with undissolved solid
c. contains as much solvent as it can hold
d. will rapidly precipitate if a seed crystal is added
e. contains no double bonds

Which one of the following is most soluble in water?

a. CH3CH2OH
b. CH3CH2CH2OH
c. CH3CH2CH2CH2OH
d. CH3OH
e. CH3CH2CH2CH2CH2OH

Which one of the following substances would be the most soluble in CCI4?

a. CH3CH2OH
b. C10H22
c. H2O
d. NaCI
e. NH3

A supersaturated solution ____________.

a. is one that has been heated
b. must be in contact with undissolved solid
c. is one with more than one solute
d. is one with a higher concentration than the solubility
e. exists only in theory and cannot actually be prepared

The molality of a benzene solution prepared by mixing 12.0 g C6H6 with 38.0 g CCl4 is _____________.

a. 0.316
b. 0.240
c. 0.622
d. 0.508
e. 4.04

Pressure has an appreciable effect on the solubility of ___________ in liquids.

a. liquids
b. salts
c. gases
d. solids
e. solids and liquids

As the concentration of a solute in a solution increases, the freezing point of the solution _________ and the vapor pressure of the solution _________.

a. increases, decreases
b. increases, increases
c. decreases, is unaffected
d. decreases, increases
e. decreases, decreases

Which liquid will have the lowest freezing point?

a. aq. 0.050 m glucose
b. aq. 0.030 m NaI
c. pure H2O
d. aq. 0.030 m AlI3
e. aq. 0.030 m CoI2

What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86°C/m.

a. -3.34
b. -1.11
c. 3.34
d. 1.11
e. 0.00

Which one of the following 0.1 M aqueous solutions would have the lowest freezing point?

a. NaCl
b. Sucrose
c. K2CrO4
d. Na2SO4
e. Al(NO3)3

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