Chemistry Test 134

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Chemistry Test 134

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Weekly Quiz Competition

Question 1
4 / -1

The heat of formation of AI₂O₃ is 380 kcals/mole and that of Fe₂O₃ is 195 kcals/mole. The heat (in kcals/mole) of the thermite reaction is

A
– 92.5

B
–185

C
–575

D
370
Question 2
4 / -1

Enthalpy of Rhombic sulphur of 25^oC is

A
–10 KJ

B
zero

C
2.2 kcal

D
–5.6 kcal

Solution

Rhombic sulphur is the standard state of sulphur or in its pure state.And in standard state it is taken as zero.
Question 3
4 / -1

Which of the following is not a state function ?

A
q/w

B
H/T

C
E, PV

D
q, w

Solution

A state function is the property of the system whose value depends only on the initial and final state of the system and is independent of the path.Heat (q) and work (W) are not state functions being path dependent.
Question 4
4 / -1

Which of the following reaction is favored by increase of temperature ?

A

B

C

D

Solution

In options a,c and d, we can say that energy is also a part of the products which is indicated by the + sign between the product and the energy but in case of reaction B, you can see that with the products energy is taken inside that is it is a negative sign so when we transfer that part of the equation to the product side, we can see that in the product there is a sign of plus along with energy which means that energy is given along with the products to obtain the reactants .you can conclude that option B is the correct answer because it is an endothermic reaction
Question 5
4 / -1

The K_p for the decomposition of SO₂CI₂ (if its degree of dissociation under one atomic pressure is 90%) is

A
2.18

B
3.16

C
4.26

D
None of these
Question 6
4 / -1

The pH of 10^-7 M HCl is

A
7

B
6.6990

C
unpredictable

D
zero

Solution

Without calculation one can also say that the answer is a because acid has a ph value less than 7.
Question 7
4 / -1

1c.c. of 0.1N HCl is added to 99 c.c. aqueous solution of sodium chloride.The pH of the resulting solution will be

A
7

B
4

C
5

D
2

Solution

Normality volume = (0.1) (0.001)

Normality volume = 0.0001 L

Volume of solution = Volume of NaCl + Volume of HCL

Volume of solution = 1 + 0.001

Volume of solution = 1.001 L

Since NaCl is a salt and is neutral so it does not effect pH of a solution.

Now,

Normality of HCL in a resulting solution = 0.0001 / 1.001

Normality of HCL in a resulting solution = 0.0001 N

So,

pH = - log [H⁺]

pH = - log (0.0001)

pH = 4

which shows that the pH of the resulting solution is 4
Question 8
4 / -1

0.2 molar aqueous solution of Ba(OH)₂ is found to be 90% ionized at 298 K.The pH of the solution is

A
10.56

B
12.56

C
13.56

D
9.56

Solution

Lets calculate the pOH first,

Since 1 mol Ba(OH)2 gives 2 moles of OH- 0.2 moles will give 0.4 mol OH-

However its 90% ionised and hence [OH-] = 0.36 M

So pOH = -log [0.36]

= 0.443

pH = 14-0.443

= 13.55
Question 9
4 / -1

If the maximum concentration of PbCI₂ in water is 0.01M at 298K, its maximum concentration in 0.1M NaCl will be

A
4 × 10^-3M

B
0.4 × 10 ^-4M

C
4 × 10 ^-2M

D
4 × 10 ^-4M

Solution

Concentrations of the ions:

The dissociation reaction for PbCl₂ in water is write as,

PbCl₂→Pb₂+(aq)+2Cl−(aq)

The maximum concentration of PbCl₂ in water is 0.01 M

We can find concentration of the ions using the stoichiometry from the above reaction.

The concentration of Pb²⁺ ion is 0.01 M and concentration of Cl⁻ ion is 2×0.01=0.02M.

Solubility product:

The solubility product of PbCl₂ can be written as,

KSP=[Pb²⁺][Cl]⁻²

Let us insert in the concentrations of Pb and Cl.

KSP=(0.01)(0.02)

Ksp =4.0×10−6

Concentration of Pb²⁺ using common ion effect:
WhenPbCl₂ is dissolved in 1MNaCl, there is a common ion Cl−. This decreases the solubility of PbCl₂ due to the common ion effect.
The concentration of Cl− in 0.1MNaCl is 0.1. Let us plug in this value to find the new concentration of Pb²⁺ ion.

Ksp=[Pb²⁺][Cl⁻]²

4×10−6=[Pb²⁺](0.1)²

[Pb²⁺]=0.014×10⁻⁶

[Pb²⁺]=0.0004M=4×10−4 M
Question 10
4 / -1

Ammonium hydrogen sulphide is contained in a closed vessel at 313 K when total pressure at equilibrium is found to be 0.8 atm. The value of K_p for the reaction

A
0.16

B
1.6

C
0.016

D
16

Solution

Ammonium hydrogen sulphide is contained in a closed vessel at 313K when total pressure at equilibrium is found to be 0.8 atm.

NH₄HS(s)⇌NH₃(g)+H₂S(g)

P_NH3=P_H2S=P

Total pressure P_NH3+P_H2S=P+P=2P

But total pressure is 0.8 atm.

2P=0.8

P=0.4 atm

P_NH3=P_H2S=P=0.4 atm

The value of the equilibrium constant for the reaction,is Kp=P_NH3×P_H2S

Kp=0.4×0.4

Kp=0.16