college level chemistry answers

7/27/2019 College level chemistry Answers

1/13

Reading graph: at 38 C the solubility of copper sulphate, CuSO4, is

28g of anhydrous salt per 100g of


2/13

Reading graph: at 84 C the solubility of potassium sulphate, K2SO4,is

22g per 100g of water.


3/13

Ex Q1: How much potassium nitrate will dissolve in 20g of water at 34C?

At 34 C the solubility is 52g per 100g of water, so scaling down, 52 x 20 / 100= 10.4g will dissolve in 20g of water.


4/13

Ex Q2: At 25 C 6.9g of copper sulphate dissolved in 30g of water,what is its solubility in g/100cm3 of water?

Scaling up, 6.9 x 100 / 30 = 23g/100g of water (check on graph, justless than 23g/100g water).


5/13

http://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppt

Henrys Law Application

The solubility of pure N2 (g) at 25oC and 1.00 atm pressure

is 6.8 x 10-4 mol/L. What is the solubility of N2 underatmospheric conditions if the partial pressure of N2 is0.78 atm?

Step 1: Use the first set of data to find k for N2 at 25C

Step 2: Use this constant to find the solubility (concentration)when P is 0.78 atm:

4

4 16.8 106.8 10

1.00

c x Mk x M atm

P atm

4 1 4(6.8 10 )(0.78 ) 5.3 10c kP x M atm atm x M
http://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppthttp://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppthttp://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppthttp://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppthttp://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppthttp://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppthttp://academic.pgcc.edu/~ssinex/Solutions.ppt


6/13

How do I get

sugar to dissolvefaster in myiced tea?

Stir, and stir, and stir

Add sugar to warm tea then add ice

Grind the sugar to a powder

Fresh solvent contact and interaction with solute

Greater surface area, more solute-solvent interaction

Faster rate of dissolution at higher temperature

http://academic.pgcc.edu/~ssinex/Solutions.ppt
http://academic.pgcc.edu/~ssinex/Solutions.ppthttp://academic.pgcc.edu/~ssinex/Solutions.ppthttp://academic.pgcc.edu/~ssinex/Solutions.ppt


7/13

Revision Units of Concentrations

amount of solute per amount of solvent or solution

Percent (by mass) =g solute

g solutionx 100

g solute

g solute + g solvent

x 100=

Molarity (M) =moles of solute

volume in liters of solution

moles = M x VL

http://academic.pgcc.edu/~ssinex/Solutions.ppt
http://academic.pgcc.edu/~ssinex/Solutions.ppthttp://academic.pgcc.edu/~ssinex/Solutions.ppt


8/13

Exampleshttp://academic.pgcc.edu/~ssinex/Solutions.ppt

What is the percent of KCl if 15 g KCl areplaced in 75 g water?

%KCl = 15g x 100/(15 g + 75 g) = 17%

What is the molarity of the KCl if 90 mL ofsolution are formed?

mole KCl = 15 g x (1 mole/74.5 g) = 0.20 mole

molarity = 0.20 mole/0.090L = 2.2 M KCl
http://academic.pgcc.edu/~ssinex/Solutions.ppthttp://academic.pgcc.edu/~ssinex/Solutions.ppt


9/13

Gas Pressure and Solubility

Quiz: The amount of dissolved oxygen in amountain lake at10,000 ft and 50oF is __?_ than theamount of dissolved oxygen in a lake nearsea level at 50oF.

Answer: Less at higher altitude becauseless pressure.

A Coke at room temperature will have __?_carbon dioxide in the gas space above the

liquid than an ice cold bottle. Answer: More gas, because the warm cokecan hold less of the gas in solution.


10/13

Gas Pressure and Solubility

Hyperbaric therapy, which involves

exposure to oxygen at higher thanatmospheric pressure may be used totreat hypoxia (low oxygen supply inthe tissues). Explain how thetreatment works.

Answer: The increase in pressure inthe chamber will cause more gases toenter into lungs.


11/13

Calculate the Freezing Point of a 4.00 molalglycol/water solution.

Kf = 1.86 oC/molal (See Kf table)

Solution

TFP = Kf m i

= (1.86 oC/molal)(4.00 m)(1)

TFP = 7.44

FP = 0 7.44 = -7.44 oC(because water normally freezes at 0)

Freezing Point Depression


12/13

At what temperature will a 5.4 molal solutionof NaCl freeze?

Solution

TFP = Kf m iTFP = (1.86 oC/molal) 5.4 m 2

TFP = 20.1oC

FP = 0 20.1 = -20.1 oC



13/13

Calculate the Freezing Point of a 4.00 molalglycol/water solution.

Kf = 1.86 oC/molal (See Kf table)

Solution

TFP = Kf m i= (1.86 oC/molal)(4.00 m)(1)

TFP = 7.44

FP = 0 7.44 = -7.44 oC(because water normally freezes at 0)


college level chemistry answers

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