Chemistry Test

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Chemistry Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

The atomic number of nitrogen is 7 and that of hydrogen is $1 .$ How many electrons are there in ammonium ion, ${N H}_{4}^{+}$ ?

A
30

B
32

C
10

D
26

Solution

When a positive charge is present on an ion then we subtract the charge from number of electrons whereas when a negative charge is present on an ion then we add the charge in total number of electrons present in the atom or compound.
Therefore, in a $N H_{4}^{+}$ ion the number of electrons present are as follows:
$= 7 + (4 \times 1) - 1$
$= 10$ electrons
Thus, we can conclude that there are 10 electrons in $N H_{4}^{+}$ ion.
Question 2
4 / -1

Which of the following would be expected to have the largest entropy per mole?

A
SO₂Cl₂(s)

B
SO₂Cl₂(g)

C
SO₂(g)

D
SO₂Cl₂(l)

Solution

If ∆H/∆T is the same for in case of SO₂ (g) and SO₂Cl₂ (g), then change in entropy is related to the molar mass of a substance. Entropies generally increase with molecular weight. Thus, SO₂Cl₂(g) has greater entropy.
Question 3
4 / -1

Acidic buffer solution is produced on mixing the aqueous solutions of

A
NH₄Cl + NH₄OH

B
CH₃COONa + CH₃COOH

C
NaCl + NaOH

D
Na₂SO₄ + H₄SO₄

Solution

Acidic buffer solution is produced on mixing the aqueous solutions of acid and acidic salt.
Question 4
4 / -1

If 0.5 g of a solute (molar mass 100 g mol^-1) in 25 g of solvent elevates the boiling point by 1 K. The molar boiling point constant of the solvent is

A
2

B
8

C
5

D
0.5

Solution

Since molality is 1/5 and molar boiling point constant is equal to 1/molality here. Hence, correct answer is 5.
Question 5
4 / -1

How many moles of magnesium phosphate Mg₃(PO₄)₂ will contain 0.25 moles of oxygen atoms?

A
2.15 x 0.02

B
3.125 x 10^-2

C
1.25 x 10^-2

D
2. 5 x 10^-2

Solution

In ${Mg}_{3} {({PO}_{4})}_{2}$ one molecule of it contains 8 atoms of oxygen
$∵ 8$ moles of oxygen atoms are present in 1 mole ${Mg}_{3} {({PO}_{4})}_{2}$
$∴ 0.25$ mole of oxygen atoms are present in
$\frac{0.25}{8} = 3.125 \times 10^{- 2} moles of {Mg}_{3} {({PO}_{4})}_{2}$
Question 6
4 / -1

The values of heat of combustion of rhombic sulphur and monoclinic sulphur are 70960 calories and 71030 calories, respectively. What is the heat of conversion of rhombic sulphur to monoclinic sulphur?

A
70960 calories

B
71030 calories

C
71570 calories

D
70 calories

Solution

ΔH (Rhombic) –ΔH (Monoclinic) = – 70960 – (– 71030) = – 70960 + 71030 = + 70 calories
Question 7
4 / -1

Which of the following aqueous solutions will have the lowest vapour pressure at room temperature?

A
0.1 M CaCl₂

B
0.1 M glucose

C
0.1 M urea

D
0.1 M NaCl

Solution

The change in the colligative properties depends on the number of particles of the non-volatile solute,
The 0.1 M solution of CaCl₂ will furnish 0.3 moles of ions (assuming α=1).
Since the particles of the non- volatile solute are maximum in CaCl₂ the change in the colligative properties will be maximum.
Therefore the solution of CaCl₂ will have the lowest vapour pressure.
0.1 mol of NaCl will furnish 0.2 moles (2 x 0.1) of ions. While urea and glucose non-ionisable. Therefore, i = 1, and both will gives 0.1 moles of ions.
Question 8
4 / -1

For Zn²⁺/Zn = - 0.76 V, Cl₂/Cl^- = 1.36 V, H⁺/ (1/2)H₂ = 0 V and Fe³⁺/Fe²⁺ = 0.77 V, the increasing order of strengths as a reducing agent is

A
Cl₂, Zn, H₂, Fe³⁺

B
H₂, Zn, Fe³⁺, Cl₂

C
Cl₂, Fe³⁺, Zn, H₂

D
Cl₂, Fe³⁺, H₂, Zn

Solution

The substances with a lower value of reduction potential are stronger reducing agents.

Hence, the increasing order of the strengths as reducing agents is:

Cl₂, Fe³⁺, H₂, Zn
Question 9
4 / -1

The plot of ln C_{0 /}C_t against t is a straight line; the order of the reaction is expected to be:

A
second order

B
first order

C
zero order

D
pseudo first order

Solution

The integrated rate law expression for the first order reaction is:
$l n \frac{C_{0}}{C_{t}} = k t$
$\ln \frac{C_{0}}{C_{t}} \propto t$
Therefore the plot between $\ln \frac{C_{0}}{C_{t}}$ and $t$ is a straight line passing through the origin. The slope of the line is given by 'k".
Question 10
4 / -1

The third line of Balmer series in the emission spectrum of hydrogen atom is due to the transition from the

A
7^th Bohr orbit to the 3^rd Bohr orbit

B
4^th Bohr orbit to the 1^st Bohr orbit

C
5^th Bohr orbit to the 2^nd Bohr orbit

D
6^th Bohr orbit to the 3^rd Bohr orbit

Solution

The first goes from 3^rd Bohr orbit to the 2^nd Bohr orbit, so the 3^rd line will be on the transition from 5^th Bohr orbit to the 2^nd Bohr orbit.

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Re-Attempt Weekly Quiz Competition

Chemistry Test - 6

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Chemistry Test - 6

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Question 1

30

32

10

26

Solution

Question 2

SO2Cl2(s)

SO2Cl2(g)

SO2(g)

SO2Cl2(l)

Solution

Question 3

NH4Cl + NH4OH

CH3COONa + CH3COOH

NaCl + NaOH

Na2SO4 + H4SO4

Solution

Question 4

2

8

5

0.5

Solution

Question 5

2.15 x 0.02

3.125 x 10-2

1.25 x 10-2

2. 5 x 10-2

Solution

Question 6

70960 calories

71030 calories

71570 calories

70 calories

Solution

Question 7

0.1 M CaCl2

0.1 M glucose

0.1 M urea

0.1 M NaCl

Solution

Question 8

Cl2, Zn, H2, Fe3+

H2, Zn, Fe3+, Cl2

Cl2, Fe3+, Zn, H2

Cl2, Fe3+, H2, Zn

Solution

Question 9

second order

first order

zero order

pseudo first order

Solution

Question 10

7th Bohr orbit to the 3rd Bohr orbit

4th Bohr orbit to the 1st Bohr orbit

5th Bohr orbit to the 2nd Bohr orbit

6th Bohr orbit to the 3rd Bohr orbit

Solution

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SO₂Cl₂(s)

SO₂Cl₂(g)

SO₂(g)

SO₂Cl₂(l)

NH₄Cl + NH₄OH

CH₃COONa + CH₃COOH

Na₂SO₄ + H₄SO₄

3.125 x 10^-2

1.25 x 10^-2

2. 5 x 10^-2

0.1 M CaCl₂

Cl₂, Zn, H₂, Fe³⁺

H₂, Zn, Fe³⁺, Cl₂

Cl₂, Fe³⁺, Zn, H₂

Cl₂, Fe³⁺, H₂, Zn

7^th Bohr orbit to the 3^rd Bohr orbit

4^th Bohr orbit to the 1^st Bohr orbit

5^th Bohr orbit to the 2^nd Bohr orbit

6^th Bohr orbit to the 3^rd Bohr orbit