Calculations in Chemistry: A Practical Approach by Kingsley Anaba

Author:Kingsley Anaba [Anaba, Kingsley]
Language: eng
Format: azw3
Published: 2015-03-13T04:00:00+00:00

Solubility Curve 7

6

5

4

3

2

1

0 0 20 40 60 80 100 Temperature oC

5. The diagram above is the solubility curve of a solute x. Find the amount of x deposited when 500cm3 of a solution of x is cooled from 60oC to 20 o C

Solution

When faced with a problem like this one, trace the solubility of the salt at 60oc and 20oc from the curve

From the curve,

The solubility of x at 60oC is 5.5 moldm-3,

at 20oC it is 4.0 moldm-3

That is to say, At 60oC, 1000cm3 of the solution contains 5.5moles And, at 20oc 1000cm3 of the solution contains 4.0moles. Solute deposited on cooling from 60oc to 20oc = 5.5 – 4 .0 = 1.5moles

On cooling from 60oc to 20oc,

1000cm3 of the solution deposits 1.5 moles

500cm3 of the solution will deposit x moles

cross multiplying and solving for x

x = 500 x 1.5

1000 = 0.75moles

The amount of solute deposited by 500cm3 is 0.75mole. 6. The solubility of sodium tetraoxosulphate (vi) is 1900g per

1000g of water at 80oC and 800g per 1000g water at 40oC. Calculate the mass of sodium tetraoxosulphate (vi) that will crystallize out of solution if 160g of the saturated solution at

80oC is cooled to 40oC.

Solution

The mass of the solution at 80oC = 1900g + 1000g = 2900g The mass of the solution at 40oC = 800g + 1000g = 1800g

Solute deposited on cooling from 80oC to 40oC = 2900 –1800 = 1100g

On cooling from 80oC to 40oC

2900g of saturated solution deposits 1100g of solute 160g of saturated solution will deposit xg of solute Cross – multiplying and solving for x x = 160 x 1100

2900 = 60. 69g

On cooling from 80oC to 40oC 160g of saturated solution will deposit 60.69g of solute.

7. If the solubility of KHCO3 is 0.40Moldm-3 at room temperature, calculate the mass of KHCO3 in 100cm3 of the solution at this temperature.

[KHCO3 = 100gmol-1]

Solution

The solubility of KHCO3 = 0.40 moldm-3 This means that,

1 dm (1000cm3) of the solution contains 0.40 mole of KHCO3

3 of the solution will contain X mole of KHCO3 100cm

Cross-multiply and solve for X

X = 100 x 0.40 1000 = 0.04 mole of KHCO3 100cm3 of the solution contains 0.04 mole of KHCO3 Now let us convert the mole to mass. Remember, Mass = Mole x Molar mass

Mass = 0.04 x 100 = 4g

Hence, 100cm3 of the solution contains 4g of KHCO3. 8. Calculate the solubility of Na2CO3 at 25 oC, if 20.0cm3 of its saturated solution at that temperature gave 1.75g of the anhydrous salt.

[C = 12, O = 16, Na = 23]

Solution

Remember, that the solubility of a solute is usually measured in mole per dm3 or gram per dm3.

If 20cm3 of the saturated solution gave 1.75g of Na2CO3, it means it contained 1.75g per 20cm3.

If 20cm3 of saturated solution contains 1.75g of Na2CO3, Then 1000cm3 of saturated solution will contain Xg of Na2CO3 Cross-multiply and solve for X

X = 1000 x 1.75

20 = 87.5g

The saturated solution contains 87.5g per dm3. Converting to mol per dm3, we apply,

moldm-3 = gram dm-3/Molar mass

molar mass of Na2CO3 = 106

moldm-3 = 87.

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