Chemistry Test 180

Result

Chemistry Test 180

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Weekly Quiz Competition

Question 1
4 / -1

The increase in volume of air, when temperature of 600 ml of it, is increased from 27°C to 47°C under constant pressure, is

A
20 mL

B
80 mL

C
40 mL

D
500 mL

Solution
Question 2
4 / -1

The density of steam at 27°C and 8.314 × 10⁴ pascal is 0.8 Kg m^–3. The compressibility factor would be

A
0.75

B
1

C
0.88

D
1.1

Solution
Question 3
4 / -1

The velocity corresponding to a gas "X" (MW = 100) corresponding to the maxima of the curve given below at a given temperature is 200 m/s. Then kinetic energy of 300 gram of gas "X" will be :

A
9000 kJ

B
9 kJ

C
90 kJ

D
810 kJ

Solution

The velocity corresponding to the maxima is the most probable velocity which is given by the expression.
Question 4
4 / -1

A 4 : 1 molar mixture of He and CH₄ gases is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The ratio of number of moles of He to CH₄ in the mixture effusing out initially will be :

A
8 : 1

B
1 : 8

C
1 : 4

D
4 : 1

Solution
Question 5
4 / -1

20 ml of H₂O₂ after acidification with dil H₂SO₄ required 30 ml of N/12 KMnO₄ for complete oxidation. The strength of H₂O₂ solution is [Molar mass of H₂O₂ = 34]

A
2 g/L

B
4 g/L

C
8 g/L

D
6 g/L

Solution
Question 6
4 / -1

3 mole of a mixture of FeSO₄ and Fe₂(SO₄)₃ required 100 ml of 2M KMnO₄ solution in acidic medium. Hence mole fraction of FeSO₄ in the mixture is

A
3/5

B
2/3

C
2/5

D
1/3

Solution
Question 7
4 / -1

Arsenic estimation can be done by Bettendorff's process. The reaction is given below :

As₄O₆ + SnCl₂ + HCl → As₄ + SnCl₄ + H₂O

Find out the exact stoichiometric ratio of the reactants : (in the order as given in question)

A
2, 4, 6

B
1, 6, 12

C
2, 8, 20

D
None of these

Solution
Question 8
4 / -1

How many moles of KMnO₄ are needed to oxidised a mixture of 1 mole of each FeSO₄ and FeC₂O₄ in acidic medium

A
4/5

B
5/4

C
3/4

D
5/3

Solution

Equivalents of KMnO₄ = equivalent of FeSO₄ + equivalent of FeC₂O₄

x = 4/5 mole
Question 9
4 / -1

The correct figure representing isothermal and adiabatic expansions of an ideal gas from a particular initial state is:

A

B

C

D

Solution

Isothermal :

pv = constant

Adiabatic :

pvγ = constant = k

Therefore,

Slope = dp / dv = - γ p / v

as γ > 1

Therefore, graph decreases more faster for adiabatic.

Stops of adiabatic curve is greater than isothermal curve.
Question 10
4 / -1

An ideal gas with C_v = 3R expands adiabatically into a vacuum thus doubling its volume. The final temperatue is given by :

A
T₂ = T₁[2^-1/3]

B
T₂ = T₁

C
T₂ = 2T₁

D

Solution

In free expansion (expansion into vacuum)

q = 0, w = 0 ⇒ ΔU = 0 , ΔT = 0

⇒ T₂ = T₁