Chemistry Test

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Chemistry Test - 19

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Weekly Quiz Competition

Question 1
4 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.

The method usually employed for dispersion of a colloidal solution is

A
addition of electrolytes

B
diffusion through animal membrane

C
condensation

D
dialysis

Solution

Peptization - Some freshly prepared precipitates, such as AgCl, Fe (OH)₃ and Al(OH)₃, can be converted into colloidal solutions by addition of small amount of suitable electrolyte. An electrolyte having an ion common with material to be dispersed is required for sol formation.

For example, the ferric hydroxide sol is obtained when small amount of ferric chloride solution is added. Similarly, Al(OH)₃ sol is obtained when small amount of hydrochloric acid is added to freshly precipitated Al(OH)₃.
Question 2
4 / -0

A mixture contains 1 mol volatile liquid \(A_{\left(P_{A}^{0}\right.}=100 \mathrm{mm} \mathrm{Hg},\) and \(3 \mathrm{mol}\) volatile liquid \(\left(\mathrm{p}_{\mathrm{g}}^{0}=80 \mathrm{mm}\right) \cdot\) If solution behaves ideally, total vapour
pressure of the solution is approximately

A
85 mm Hg

B
86 mm Hg

C
90 mm Hg

D
92 mm Hg

Solution

Mole fraction of A is \(X_{A}=\frac{1}{1+3}=0.25\)
The mole fraction of \(\mathrm{B}\) is \(X_{B}=1-0.25=0.75\)
The expression for the total pressure of the solution is
\(P=P_{A}^{o} X_{A}+P_{B}^{o} X_{B}\)
Substitute values in the above expression
\(P=100 \times 0.25+80 \times 0.75=85 \mathrm{mm} \mathrm{Hg}\)
Hence, the correct option is \(A\)
Question 3
4 / -0

How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00 g? [Atomic masses : Na = 23 u, Cl = 35.5 u]

A
5.14 *10²¹

B
2.57 * 10²¹

C
1.28 * 10²¹

D
1.71 *10²¹

Solution

\[
\begin{aligned}
\text { Mass of } 1 \text { unit cell }=& \text { volume } \times \text { density } \\
&=a^{2} \times d \\
&=\frac{a^{3} \times M \times Z}{N^{0} a^{0}} \\
&=\frac{58.5 \times 4}{6.023 \times 10^{23}}
\end{aligned}
\]
Number of unit cells in \(1 \mathrm{gm}=1 / \mathrm{M}\)
\[
\begin{array}{l}
=6.023 \times 10^{23} / 58.5 \times 4 \\
=2.57 \times 10^{21}
\end{array}
\]
Question 4
4 / -0

The energies of activation for forward and reverse reactions of A2 + B2 → 2AB are 180 kJ mol-1 and 200 kJ mol^-1 respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol^-1. The enthalpy change of the reaction (A2 + B2 → 2AB) in the presence of catalyst will be (in kJ mol^-1)

A
280

B
-20

C
300

D
-120

Solution

ΔH does not change due to catalyst.

ΔH = 180 -200 = -20
Question 5
4 / -0

The minimum amount of iron required to produce just sufficient hydrogen (by reaction with dil. HCl) to react completely with the oxygen obtained by heating 2.45 g KClO₃ is

A
2.52 g

B
3.36 g

C
3.00 g

D
2.50 g

Solution

\(2 \mathrm{KClO}_{3} \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_{2}\)
\(\therefore 245 \mathrm{g} \mathrm{KClO}_{3}\) produces 3 moles oxygen or \(2.45 \mathrm{g} \mathrm{KClO}_{3}\) gives \(0.03 \mathrm{mole}\) oxygen.
\(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) or 1 mole oxygen reacts with 2 moles hydrogen or 0.03 mole oxygen will
react with 0.06 mole hydrogen.
\(\mathrm{Fe}+2 \mathrm{HCl} \rightarrow \mathrm{FeCl}_{2}+\mathrm{H}_{2}\) or 1 mole hydrogen is obtained from \(56 \mathrm{g}\) Fe or \(0.06 \mathrm{mole}\) hydrogen
will be obtained from \(56 \times 0.06 \mathrm{g} \mathrm{Fe}=3.36 \mathrm{g} \mathrm{Fe}\)
The solution can also be obtained by the equivalent concept:
\((G . E q)_{K C l O_{3}}=(G, E q)_{F e}\)
\((n \times N-f a c t o r)_{K C l O_{3}}=(n \times N-f a c t o r)_{F e}\)
\(\frac{2.45}{122.5} \times 6=n \times 2\)
n Fe required \(=0.06\)
Mass of Iron required \(=0.06 \times 56.0=3.36 \mathrm{g}\)
Question 6
4 / -0

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), H is equal to

A
ΔE + 2RT

B
ΔE - 2RT

C
ΔE + RT

D
ΔE - RT

Solution

We know that ΔH = ΔE + ΔnRT

In the present case, Δn = - 2

ΔH = ΔE - 2RT
Question 7
4 / -0

In adsorption, rate of physisorption increases when

A
temperature is decreased

B
temperature is increased

C
pressure is decreased

D
None of these

Solution

The rate of physical adsorption increases on decreasing the temperature and the rate decreases on increasing the temperature.
Question 8
4 / -0

The gases which diffuse at the same rate under the same temperature and pressure conditions are

A
CO and NO

B
NO₂ and CO₂

C
NH₃ and PH₃

D
NO and C₂H₆

Solution

According to Graham's law of diffusion, \(\frac{r_{1}}{r_{2}} \alpha \sqrt{\frac{M_{2}}{M_{1}}}\)
since molecular masses of both NO and \(\mathrm{C}_{2} \mathrm{H}_{6}\) are the same, so their rates of diffusion will also be the
same under the same temperature and pressure conditions.
Question 9
4 / -0

Directions: In the following question, a statement of Assertion is given; followed by a corresponding statement of Reason just below it. Of the statement mark the correct answer as:

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true, but the reason is not the correct explanation of the assertion.

(c) If assertion is true statement, but reason is false.

(d) If both assertion and reason are false.

Assertion: Molar entropy of vaporisation of water is different from ethanol.

Reason: Water is more polar than ethanol.

A
(a)

B
(b)

C
(c)

D
(d)

Solution

Molar entropy of vaporisation of water is different from ethanol. Molar entropy is independent of polarity of the molecule.
Question 10
4 / -0

In which of the following processes, maximum increase in entropy is observed?

A
Melting of ice

B
Sublimation of naphthalene

C
Condensation of water

D
Condensation of water

Solution

Entropy of gaseous phase is maximum. So sublimation of naphthalene will involve maximum entropy increase.

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Chemistry Test - 19

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Chemistry Test - 19

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Question 1

addition of electrolytes

diffusion through animal membrane

condensation

dialysis

Solution

Question 2

85 mm Hg

86 mm Hg

90 mm Hg

92 mm Hg

Solution

Question 3

5.14 *1021

2.57 * 1021

1.28 * 1021

1.71 *1021

Solution

Question 4

280

-20

300

-120

Solution

Question 5

2.52 g

3.36 g

3.00 g

2.50 g

Solution

Question 6

ΔE + 2RT

ΔE - 2RT

ΔE + RT

ΔE - RT

Solution

Question 7

temperature is decreased

temperature is increased

pressure is decreased

None of these

Solution

Question 8

CO and NO

NO2 and CO2

NH3 and PH3

NO and C2H6

Solution

Question 9

(a)

(b)

(c)

(d)

Solution

Question 10

Melting of ice

Sublimation of naphthalene

Condensation of water

Condensation of water

Solution

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5.14 *10²¹

2.57 * 10²¹

1.28 * 10²¹

1.71 *10²¹

NO₂ and CO₂

NH₃ and PH₃

NO and C₂H₆