Question 5 Generally, for solubility of a gas, as the temperature_________ the s

Question

Question 5
Generally, for solubility of a gas, as the temperature_________ the solubility _________.
a) decreases, does not change
b) increases, increases
c) increases, decreases
d) decreases, decreases

Question 6
A common laboratory technique for degassing a solvent is to place it in a flask that is sealed to the atmosphere and then evacuate the flask to remove any gases above the liquid. Why is this procedure effective?
a) Increasing the pressure of a gas above the liquid decreases it's solubility
b) solubility of gas is independent of it's pressure
c) decreasing the pressure of a gas above the liquid decreases it's solubility
d )decreasing the pressure of a gas above the liquid increases it's solubility

Question 7
Which of the solution has the highest freezing point?
a) 0.25 m CH3OH
b) 0.05 m K2SO4
c) 0.1 m NaCl
d) 0.1 m MgBr2

Question 8
What is the molar solubility (in M) of helium at 1.0 atm and 25 oC?
a) 0.037
b) 0.00037
c) 0.37
d) 0.0037

Question 9
Which compound has the highest solubility in water?
a) O2
b) NH3
c) CH4
d) CO2

Explanation / Answer

Answer – 5) We know the solubility of a gas is directly proportional with temperature, so

Generally, for solubility of a gas, as the temperature increases the solubility increases.

6) We know the pressure is directly proportional with solubility for the gases, so there is answer is -

c) decreasing the pressure of a gas above the liquid decreases it's solubility.

Q 7) We know the freezing point is depending on the freezing point depression.

Tf = i*Kf*m

So as the molality and Van’t off factor increase there is freezing point depression high and the freezing point gets higher.

So there is first CH3OH i = 1 , for K2SO4 , i = 3

NaCl i = 2 and for MgBr2 there are 3.

So for the d) 0.1 m MgBr2 highest freezing point.

Q 8) Given, P = 1.0 atm , T = 25+273 = 298 K , moles = 1 moles

We know the solubility is moles/L

Ideal gas law

PV = nRT

So, n/V = P /RT

              = 1.0 atm / 0.0821 L.atm.mol-1.K-1 * 298 K

           = 0.040 moles/L

So answer is a) 0.037 mole/L

Q 9) We know water is the polar molecule and we also know the like dissolve in like so there is O2 CH4 and CO2 are non polar, so they are not soluble in the water.

NH3 is polar, so it is dissolve highly in the water, so answer for this one is b) NH3

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