Topic: UNIT 9: Solutions (Test 1)



Topic: UNIT 9: Solutions
Q.1
The molality of a solution containing 1.8g of glucose dissolved in 250g of water is
A. 0.2 M
B. 0.01 M
C. 0.02 M
D. 0.04 M
Answer : Option D
Explaination / Solution:

molality = number of moles of solute / weight of solvent (in kg)

= (1.8/180) / 0.25 = 0.01/0.25 = 0.04M


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Q.2
Which of the following concentration terms is / are independent of temperature
A. molality
B. molarity
C. mole fraction
D. (a) and (c)
Answer : Option D
Explaination / Solution:

Molality and mole fraction are independent of temperature.

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Q.3

Stomach acid, a dilute solution of HCl can be neutralised by reaction with Aluminium hydroxide

Al (OH)3 + 3HCl (aq) → AlCl3 + 3 H2O

How many millilitres of 0.1 M Al(OH)3 solution are needed to neutralise 21 mL of 0.1 M HCl ?

A. 14 mL
B. 7 mL
C. 21 mL
D. none of these
Answer : Option B
Explaination / Solution:

M1 × V1 = M2 × V2                [ 0.1 M

Al(OH)3 gives 3 × 0.1 = 0.3 M OH– ions]

0.3 × V1 = 0.1 × 21

V1 = 01x21 / 0.3 = 7ml


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Q.4
The partial pressure of nitrogen in air is 0.76 atm and its Henry's law constant is 7.6 × 104 atm at 300K. What is the molefraction of nitrogen gas in the solution obtained when air is bubbled through water at 300K ?
A. 1 × 10–4 
B. 1 × 10–6
C. 2 × 10–5
D. 1 × 10–5
Answer : Option D
Explaination / Solution:

PN2= 0.76 atm

KH = 7.6 × 104

x = ?

PN2 = KH . x

0.76 = 7.6 × 104 × x

so x = 0.76 / [7.6 x104] = 1x10-5


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Q.5
The Henry's law constant for the solubility of Nitrogen gas in water at 350 K is 8 × 104 atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350K and 4 atm pressure is
A. 4 × 10–4
B. 4 × 104
C. 2 × 10–2
D. 2.5 × 10–4
Answer : Option D
Explaination / Solution:

KH = 8 × 104

(xN2)in air = 0.5

Total pressure = 4 atm

pressure Partial pressure of nitrogen = mole fraction × total pressure = 0.5 × 4 = 2



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Q.6
Which one of the following is incorrect for ideal solution ?
A. ∆Hmix = 0
B. ∆Umix = 0
C. ∆P = Pobserved – PCalculated by raoults law = 0
D. ∆Gmix = 0
Answer : Option D
Explaination / Solution:

for an ideal solution,

ΔSmix ≠ 0 ; Hence ΔGmix ≠ 0

incorrect is ΔGmix = 0


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Q.7
Which one of the following gases has the lowest value of Henry's law constant ?
A. N2
B. He
C. CO2
D. H2
Answer : Option C
Explaination / Solution:

Carbon dioxide ; most stable gas and has lowest value of Henrys Law constant.

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Q.8
P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be
A. P1 + x1 (P2 – P1)
B. P2 – x1 (P2 + P1)
C. P1 – x2 (P1 – P2)
D. P1 + x2 (P1 – P2)
Answer : Option C
Explaination / Solution:

Ptotal = P1 + P2

= P1x1 + P2x2           (x1 + x2 = 1)

= P1 (1 – x2) + P2x2    (x1 = 1 – x2)

= P1 – P1x2 + P2x2

= P1 – x2 (P1 – P2)


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Q.9
Osometic pressure (p) of a solution is given by the relation
A. π = nRT
B. πV = nRT
C. πRT = n
D. none of these
Answer : Option B
Explaination / Solution:

π = CRT

π = n/V . RT

π V = nRT 


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Q.10
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law ?
A. Acetone + chloroform
B. Water + nitric acid
C. HCl + water
D. ethanol + water
Answer : Option D
Explaination / Solution:

Ethanol and water

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