1 mark questions and answers - Choose the best answer: Chemistry: Solutions

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**

**Solution:**

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

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

2.
Which of the following concentration terms is / are independent of temperature

a)
molality

b)
molarity

c)
mole fraction

**d) (a) and (c)**

**Solution:**

option
(d) is correct. Molality and mole fraction are independent of temperature.

3.
Stomach acid, a dilute solution of HCl can be neutralised by reaction with
Aluminium hydroxide

Al
(OH)_{3} + 3HCl (aq) → AlCl_{3} + 3 H_{2}O

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

**Solution:**

M_{1}
× V_{1} = M_{2} × V_{2} [∵ 0.1 M

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

0.3
× V_{1} = 0.1 × 21

V_{1}
= 01x21 / 0.3 = 7ml

4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry's law
constant is 7.6 × 10^{4} 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}**

**Solution:**

P_{N2}=
0.76 atm

K_{H}
= 7.6 × 10^{4}

x
= ?

P_{N2}
= K_{H} . x

0.76
= 7.6 × 10^{4} × x

so
x = 0.76 / [7.6 x10^{4}] = 1x10^{-5}

5.
The Henry's law constant for the solubility of Nitrogen gas in water at 350 K
is 8 × 10^{4} 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 × 10^{4}

c)
2 × 10^{–2}

**d) 2.5 × 10 ^{–4}**

**Solution:**

K_{H}
= 8 × 10^{4}

(x_{N2})_{in
air} = 0.5

Total
pressure = 4 atm

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

6.
Which one of the following is incorrect for ideal solution ?

a)
∆H_{mix} = 0

b)
∆U_{mix} = 0

c)
∆P = P_{observed} – P_{Calculated by raoults law} = 0

**d) ∆G _{mix} = 0**

**Solution:**

for
an ideal solution,

ΔS_{mix}
≠ 0 ; Hence ΔG_{mix} ≠ 0

∴ incorrect is ΔG_{mix} = 0

7.
Which one of the following gases has the lowest value of Henry's law constant ?

a)
N_{2}

b)
He

**c) CO _{2}**

d)
H_{2}

**Solution:**

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

8.
P_{1} and P_{2} are the vapour pressures of pure liquid
components, 1 and 2 respectively of an ideal binary solution if x_{1}
represents the mole fraction of component 1, the total pressure of the solution
formed by 1 and 2 will be

a)
P_{1} + x_{1} (P_{2} – P_{1})

b)
P_{2} – x_{1} (P_{2} + P_{1})

**c) P _{1} – x_{2} (P_{1}
– P_{2})**

d)
P_{1} + x_{2} (P_{1} – P_{2})

**Solution:**

P_{total}
= P_{1} + P_{2}

=
P_{1}x_{1} + P_{2}x_{2} (x_{1} + x_{2} = 1)

=
P_{1} (1 – x_{2}) + P_{2}x_{2} (x_{1} = 1 – x_{2})

=
P_{1} – P_{1}x_{2} + P_{2}x_{2}

=
P_{1} – x_{2} (P_{1} – P_{2})

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

**Solution:**

π
= CRT

π
= n/V . RT

π
V = nRT

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**

**Solution:**

Ethanol
and water

11.
The Henry's law constants for two gases A and B are x and y respectively. The
ratio of mole fractions of A to B is 0.2. The ratio of mole fraction of B and A
dissolved in water will be

a)
2x/y

b)
y/0.2x

c)
0.2x/y

**d) 5x/y**

**Solution:**

Given,
(K_{H})_{A} = x

(K_{H})_{B}
= y

x_{A}/x_{B}
= 0.2

(x_{A}/x_{B})_{in
solution} = ?

P_{A}
= x (x_{A}) _{in solution} – (1)

P_{B}
= y (x_{B}) _{in solution} – (2)

(x_{A}/x_{B})_{in
solution} = (P_{B}/P_{A}) . (x/y) = (x_{B}/x_{A})
. (x/y) = (1/0.2) . (x/y) = 5x/y

12.
At 100^{o}C the vapour pressure of a solution containing 6.5g a solute
in 100g water is 732mm. If K_{b} = 0.52, the boiling point of this solution
will be

a)
102^{o}C

b)
100^{o}C

**c) 101 ^{o}C**

d)
100.52^{o}C

**Solution:**

T_{b}
– 100 = 1.06

T_{b}
= 100 + 1.06

=
101.06 ≈ 101o C

13.
According to Raoults law, the relative lowering of vapour pressure for a
solution is equal to

a)
mole fraction of solvent

**b) mole fraction of solute**

c)
number of moles of solute

d)
number of moles of solvent

**Solution:**

∆P/Pº=
x_{2} (mole fraction of the solute)

14.
At same temperature, which pair of the following solutions are isotonic ?

a)
0.2 M BaCl_{2} and 0.2M urea

b)
0.1 M glucose and 0.2 M urea

c)
0.1 M NaCl and 0.1 M K_{2}SO_{4}

**d) 0.1 M Ba (NO _{3})_{2}
and 0.1 M Na_{2} SO_{4}**

**Solution:**

0.1 × 3 ion [Ba^{2+}, 2NO_{3}^{–}]

= 0.1 × 3 ion [2 Na^{+}, SO_{4}^{–}]

15.
The empirical formula of a non-electrolyte(X) is CH_{2}O. A solution
containing six gram of X exerts the same osmotic pressure as that of 0.025M
glucose solution at the same temperature. The molecular formula of X is

a)
C_{2}H_{4}O_{2}

**b) C _{8}H_{16}O_{8}**

c)
C_{4}H_{8}O_{4}

d)
CH_{2}O

**Solution:**

(π_{1})_{non
electrolyte} = (π_{2})_{glucose}

C_{1}RT
= C_{2}RT

[CH_{2}O => 12+2+16=30]

W_{1}/M_{1}
= W_{1}/M_{2}

6 / n(30) = 0.025

n =
8

∴ molecular formula = C_{8}H_{16}O_{8}

16.
The K_{H} for the solution of oxygen dissolved in water is 4 × 10^{4}
atm at a given temperature. If the partial pressure of oxygen in air is 0.4
atm, the mole fraction of oxygen in solution is

a)
4.6 × 10^{3}

b)
1.6 × 10^{4}

**c) 1 × 10 ^{–5}**

d)
1 × 10^{5}

**Solution:**

17.
Normality of 1.25M sulphuric acid is

a)
1.25 N

b)
3.75 N

**c) 2.5 N**

d)
2.25 N

**Solution:**

Normality
of H_{2}SO_{4} = (no.of replacable H^{+})×M

=
2 × 1.25

=
2.5 N

18.
Two liquids X and Y on mixing gives a warm solution. The solution is

a)
ideal

b)
non-ideal and shows positive deviation from Raoults law

c)
ideal and shows negative deviation from Raoults Law

**d) non-ideal and shows negative
deviation from Raoults Law**

**Solution:**

ΔH_{mix}
is negative and show negative deviation from Raoults law.

19.
The relative lowering of vapour pressure of a sugar solution in water is 3.5 × 10^{–3}.
The mole fraction of water in that solution is

a)
0.0035

b)
0.35

c)
0.0035 / 18

**d) 0.9965**

**Solution:**

∆_{P }/ Pº = X_{sugar}

3.5
× 10^{–3} = X_{sugar}

X_{sugar}
+ X_{H}_{2}_{o} = 1

X_{H}_{2}_{o} = 1 – 0.0035 = 0.9965

20.
The mass of a non-voltaile solute (molar mass 80 g mol^{–1}) which
should be dissolved in 92g of toluene to reduce its vapour pressure to 90%

a)
10g

b)
20g

c)
9.2 g

**d) 8.89g**

**Solution:**

21.
For a solution, the plot of osmotic pressure (p) verses the concentration (c in
mol L^{–1}) gives a straight line with slope 310R where 'R' is the gas
constant. The temperature at which osmotic pressure measured is

a)
310 × 0.082 K

b)
310 º C

**c) 37 ^{o}C **

d)
310/0.082 K

**Solution:**

π
= CRT

y
= x (m)

m
= RT

310
R = RT

∴ T = 310 K

=
37º C

22.
200ml of an aqueous solution of a protein contains 1.26g of protein. At 300K,
the osmotic pressure of this solution is found to be 2.52 × 10^{–3}
bar. The molar mass of protein will be (R = 0.083 L bar mol^{–1}K^{–1})

**a) 62.22 Kg mol ^{–1} **

b)
12444g mol^{–1}

c)
300g mol^{–1}

d)
none of these

**Solution:**

π
= CRT

π
= [ W/MV ] . RT

M
= WRT / π V

=
[1.26 x 0.083x300] / [2.52x10^{-3}x0.2]

=
62.22 Kg mol^{-1}

23.
The Van't Hoff factor (i) for a dilute aqueous solution of the strong
elecrolyte barium hydroxide is

a)
0

b)
1

c)
2

**d) 3**

**Solution:**

Ba
(OH)_{2} dissociates to form

Ba^{2+}
and 2OH ion

α=(i-1) / (n-1)

i=
α(n-1)+1

n=i=3 (for Ba(OH)_{2}), α =1)

24.
What is the molality of a 10% W/W aqueous sodium hydroxide solution ?

a)
2.778

**b) 2.5**

c)
10

d)
0.4

**Solution:**

10%
W/W aqueous NaOH solution means that 10 g of sodium hydroxide in 100g solution

Molality
= no. of moles of solute / weight of solvent (in kg)

=
(10/40) / 0.1 = 0.25/0.1 = 2.5M

25.
The correct equation for the degree of an associating solute, 'n' molecules of
which undergoes association in solution, is

a)
α = n(i −1)/ (n −1)

b)
α^{2} = n(1 − i) / n −1

**c) α = n(i −1)/(1-i) **

d)
α = n(1 − n) / n(1-i)

**Solution:**

α=(1-i)n/(n-1) or n(i-1)/(1-n)

26.
Which of the following aqueous solutions has the highest boiling point ?

a)
0.1 M KNO_{3}

**b) 0.1 M Na _{3}PO_{4}**

c)
0.1 M BaCl_{2}

d)
0.1 M K_{2}SO_{4}

**Solution:**

Elevation
of boiling point is more in the case of Na_{3}PO_{4} (no. of
ions 4 ; 3 Na^{+}, PO_{4}^{3–})

27.
The freezing point depression constant for water is 1.86^{o} K Kgmol^{-1}.
If 5g Na_{2}SO_{4} is dissolved in 45g water, the depression in
freezing point is 3.64oC. The Vant Hoff factor for Na_{2}SO_{4}
is

**a) 2.50**

b)
2.63

c)
3.64

d)
5.50

**Solution:**

28. Equimolal aqueous solutions of NaCl and KCl are prepared. If the freezing point of NaCl is –2ºC, the freezing point of KCl solution is expected to be

**a) –2 ^{o}C**

b)
– 4^{o}C

c)
– 1^{o}C

d)
0^{o}C

**Solution:**

Equimolal
aqueous solution of KCl also shows 2ºC depression in freezing point.

29.
Phenol dimerises in benzene having van't Hoff factor 0.54. What is the degree
of association ?

a)
0.46

b)
92

c)
46

**d) 0.92**

**Solution:**

i
= 0.54

α=(1-i)n/(n-1)

=(1-0.54)2
/ (2-1)

α= 0.92

30.
Assertion : An ideal solution obeys Raoults Law

Reason
: In an ideal solution, solvent-solvent as well as solute-solute interactions
are similar to solute-solvent interactions.

**a) both assertion and reason are
true and reason is the correct explanation of assertion**

b)
both assertion and reason are true but reason is not the correct explanation of
assertion

c)
assertion is true but reason is false

d)
both assertion and reason are false

**Solution:**

both
assertion and reason are correct and reason is the correct explanation of
assertion.

Tags : 11th Chemistry : Solutions

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