Chemistry Test

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Chemistry Test - 18

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Question 1/10
1 / -0

The most acidic oxyacid of halogen among the given option is

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A
HIO

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B
H₅IO₆

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C
HIO₃

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D
HF

Solutions

The acidity of oxyacids of halogen can be compared by checking the oxidation state of halogen in the acid. Higher the oxidation state, higher the acidity. Here H₅IO₆ has an oxidation state of +7 and hence is the strongest acid.
Question 2/10
1 / -0

50cm³ of 0.04MK₂Cr₂O₇ in acidic medium oxidizes a sample of H₂S gas to sulphur. Volume of 0.03MKMnO₄ required to oxidize the same amount of H₂S gas to sulphur, in acidic medium is :

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A
80cm³

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B
120cm³

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C
60cm³

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D
90cm³

Solutions
Question 3/10
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Excess of KI reacts with CuSO₄ solution and then Na₂S₂O₃ solution is added to it. Which of the following statements is incorrect for these reactions?

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A
Na₂S₂O₃is oxidised

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B
CuI₂ is formed

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C
Cu₂I₂ is formed

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D
Evolved I₂ is reduced

Solutions

The reactions that take place when excess of $K I$ reacts with $C u S O_{4}$ and then adding $N a_{2} S_{2} O_{3}$ are shown below:
$4 K I + 2 C u S O_{4} \to I_{2} + C u_{2} I_{2} + 2 K_{2} S O_{4}$
$I_{2} + 2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + 2 N a I$
During this process, $C u_{2} I_{2}$ is formed while $C u I_{2}$ is not formed. $N a_{2} S_{2} O_{3}$ reduces the liberated iodine. In the second reaction, the oxidation state of iodine changes from 0 to $- 1,$ while that of sulphur changes from +2 to +2.5
Question 4/10
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A two litre flask contains 22 g of carbon dioxide and 1 g of helium at 20⁰ C. Calculate the partial pressure exerted by CO₂ and He if the total pressure is 3 atm.

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A
2.25 atm, 0.75 atm

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B
1.5 atm, 1.5 atm

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C
2.5 atm, 0.5 atm

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D
2 atm, 1 atm

Solutions

The molar masses of carbon dioxide and helium are $44 g / mol$ and $4 g / mol$ respectively. The number of moles is the ratio of mass to molar mass.
$22 g$ of carbon dioxide $= \frac{22}{44} = 0.5 moles$
$1 g$ of helium $= \frac{1}{4} = 0.25 moles$
The mole fraction of carbon dioxide $= \frac{0.5}{0.5 + 0.25} = 0.667$
The mole fraction of helium $= \frac{0.25}{0.5 + 0.25} = 0.333$
The partial pressure of a gas is the product of its mole fraction and total pressure. The partial pressure of carbon dioxide $= 0.667 \times 3 = 2 atm$
The partial pressure of helium $= 0.333 \times 3 = 1$ atm.
Question 5/10
1 / -0

3.6 gm of an ideal gas was injected into a bulb of internal volume of 8 L at pressure P atm and temp T − K. The bulb was then placed in a thermostat maintained at (T+15) K.0.6 gm of the gas was

let off to keep the original pressure. Find P and T if mol weight of gas is 44.

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A
P=0.062 atm, T=75

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B
P=1.062 atm, T=75

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C
P=2.062 atm, T=55 K

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D
P=3.062 atm, T=55 K

Solutions

$n = \frac{3.6}{44} = 0.0818$
$n^{'} = \frac{3.6 - 0.6}{44} = 0.06818$
since $P$ and $V$ are constant, $n T = n^{'} T^{'}$ $0.0818 T = 0.06818 T + 1.0227$
$T = 75 K$
$P = \frac{n R T}{V} = \frac{0.0818 \times 75 \times 0.0821}{8} = 0.062 atm$
Question 6/10
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Children brought some balloons each of 2 litre capacity to a chemist and asked him to fill them with hydrogen gas. The chemist possessed 8 litre cylinder containing hydrogen at 10 atm pressure at room temperature. How many balloons could he fill with hydrogen gas at normal atmospheric pressure at the same temperature?

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A
40

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B
36

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C
32

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D
28

Solutions

Initial pressure in the cylinder $P = 10$ atm. Initial volume in the cylinder $V = 8 L$ . Final pressure in the balloons $P^{'} = 1$ atm. Final volume in the balloons $V^{'} = ? ?$ $P V = P^{'} V^{'}$ at same temperature 10 atm $\times 8 L = 1$ atm $\times V^{'}$
$V^{'} = 80 L$
The number of balloons ( each of 2 litre capacity ) that he could fill
$= \frac{80 L}{2 L} = 40$
Question 7/10
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While resting, the average human male use 0.2 dm³ of O₂ per hour at 1 atm & 273 K for each kg of body mass. Assume that all this O₂ is used to produce energy by oxidising glucose in the body. What is the mass of glucose required per hour by a resting male having mass 60 kg. What volume, at 1 atm and 273 K of CO₂ would be produced?

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A
16.07 gm ; 12 dm³

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B
26.07 gm ; 12 dm³

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C
16.07 gm ; 22 dm³

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D
26.07 gm ; 22 dm³

Solutions

$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$
$n_{O_{2}} = 60 \times \frac{0.2}{22.4} = 0.5357$
$n_{g l u c o s e} = \frac{0.5357}{6} = 0.08928$
$W_{g l u c o s e} = 180 \times 0.08928 = 16.07$
$n_{C O_{2}} = n_{O_{2}} = 0.5357$
$V_{C O_{2}} = 22.4 \times 0.5357 = 12 d m^{3}$
Question 8/10
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The mass of potassium dichromate crystals required to oixidise 750 cm³ of 0.6 M Mohr's salt solution is : (Given, molar mass, potassium dichromate = 294, Mohr's salt = 392)

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0.39 g

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B
0.37 g

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C
22.05 g

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D
2.2 g

Solutions

Amount of Mohr's salt oxidized is $0.75 mL \times 0.6 mol / l = 0.45 mol$
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} {(S O_{4})}_{3} + 3 F e_{2} {(S O_{4})}_{3} + 7 H_{2} O$
For 6 mol of Mohr's salt 1 mol of Pottasium dichromate is required For 0.45 mol of Mohr's salt, Pottasium dichromate required is $\frac{1 \times 0.45}{6} = 0.075 mol$ Mass of potassium dichromate required is $0.075 \times 294 = 22.05 g$
Question 9/10
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At a certain temperature the time required for the complete diffusion of 200 ml of H₂ gas is 30 min. The time required for the complete diffusion of 50 ml of O₂ gas at the same temperature will be:

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A
60 min

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B
30 min

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C
45 min

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D
15 min

Solutions

According to Graham's law of diffusion,
Rate of diffusion, $Γ \propto \frac{1}{\sqrt{M}} = \frac{V}{t}$
[Here, $M =$ molecular mass, $V =$ volume and $t =$ time $]$ Thus, for $H_{2}$ gas,
$\frac{200}{30} = \frac{1}{\sqrt{2}}$
For $O_{2}$ gas,
$\frac{50}{t} = \frac{1}{\sqrt{32}}$
$\dots \dots . . ($ ii $)$
From equations (i) and (ii), we get
$\frac{200 \times t}{30 \times 50} = \frac{\sqrt{32}}{\sqrt{2}}$
$t = \frac{\sqrt{16} \times 30 \times 50}{200}$
$= \frac{4 \times 30}{4} = 30 \min$
Question 10/10
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If 20 mL of 0.1 M K₂Cr₂O₇ is required to titrate 10 mL of a liquid iron supplement, then the concentration of iron in the the the vitamin solution is :

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1.2 M

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B
2.4 M

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C
0.6 N

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D
1.56 M

Solutions

The redox reaction is as follows:
$C r_{2} O_{7}^{2 -} + F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + F e^{3 +} + 7 H_{2} O$
The redox changes involved are given below:
$i .6 e^{-} + C r_{2} O_{7}^{2 -} \to 2 C r^{3 +} (x = 6)$
ii. $F e^{2 +} \to F e^{3 +} + e^{-} (x = 1)$
Milliequivalents of ${Cr}_{2} O_{7}^{2 -} =$ Milliequivalents of ${Fe}^{2 +}$ $20 \times 0.1 \times 6 = 10 \times N$
$N = 1.2$
$M = \frac{N}{^{'} n^{'} f a c t o r} = \frac{1.2}{1} = 1.2 M$
Hence, the concentration of iron in the vitamin solution is $1.2 M$ .

Question 1/10

HIO

H5IO6

HIO3

HF

Question 2/10

80cm3

120cm3

60cm3

90cm3

Question 3/10

Na2S2O3is oxidised

CuI2 is formed

Cu2I2 is formed

Evolved I2 is reduced

Question 4/10

2.25 atm, 0.75 atm

1.5 atm, 1.5 atm

2.5 atm, 0.5 atm

2 atm, 1 atm

Question 5/10

P=0.062 atm, T=75

P=1.062 atm, T=75

P=2.062 atm, T=55 K

P=3.062 atm, T=55 K

Question 6/10

40

36

32

28

Question 7/10

16.07 gm ; 12 dm3

26.07 gm ; 12 dm3

16.07 gm ; 22 dm3

26.07 gm ; 22 dm3

Question 8/10

0.39 g

0.37 g

22.05 g

2.2 g

Question 9/10

60 min

30 min

45 min

15 min

Question 10/10

1.2 M

2.4 M

0.6 N

1.56 M

H₅IO₆

HIO₃

80cm³

120cm³

60cm³

90cm³

Na₂S₂O₃is oxidised

CuI₂ is formed

Cu₂I₂ is formed

Evolved I₂ is reduced

16.07 gm ; 12 dm³

26.07 gm ; 12 dm³

16.07 gm ; 22 dm³

26.07 gm ; 22 dm³