Chemistry Test

Result

Chemistry Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Directions: The following question has four choices, out of which only one is correct.

Hydroxyl amine reduces iron(III) according to the following equation.

NH₂OH + Fe₂(SO₄)₃ → N₂ + H₂O + FeSO₄ + H₂SO₄

Which of the following statements is/are correct?

A
The acidity of hydroxyl amine is 1.

B
Equivalent weight of Fe₂(SO₄)₃ is M/2.

C
6 milliequivalent of Fe₂(SO₄)₃ is contained in 3 millimoles of ferric sulphate.

D
All of the above

Solution

Hydroxyl amine is a Lewis base and acidity of a Lewis base is given by the number of donatable electron pairs, which is one in this case. Hence, option (1) is correct.

2Fe⁺² → Fe⁺³. The change in oxidation state is 2 x 1 = 2. Hence, option (2) is correct.

Milliequivalent = Millimoles x n-factor.

Since n-factor is 2 in this case, so this statement is correct.

Hence, the correct answer is option (D).
Question 2
1 / -0

Consider the reaction between potassium permanganate and a metal sulphite represented as follows.
\(Mn O _{4}^{-}+ SO _{3}^{2-}+ H ^{+} \rightarrow Mn ^{2+}+ S O _{4}^{2-}+ H _{2} O\)
Which of the following options about the process is correct?

A
1 mole of potassium permanganate oxidises 5 moles of the metal sulphite.

B
There is a net transfer of 5 electrons in the reaction.

C
20 mL of 0.5M KMnO₄ solution can neutralise 10 mL of 0.25 M metal sulphite.

D
50 mL of KMnO₄ can oxidise a 20 mL equimolar solution of the metal sulphite.

Solution

\(m . E q\) of \(K M n O_{4}=m .\) Eq of metal sulphite
Using \(m . Eq = M \times v \times N\) -factor
\((0.5 \times v \times 5)_{K M n O_{4}}=(0.25 \times 10 \times 2)_{\text {metal sulphite }}\)
\(v=20 m L\)
\(2 Mn O _{4}^{-}+5 S O _{3}^{2-}+6 H ^{+} \rightarrow 2 Mn ^{2+}+5 S O _{4}^{2-}+3 H _{2} O\)
The balanced equation shows that 1 mole of potassium permanganate will oxidise 2.5 moles of the metal sulphite.
There is a net transfer of 10 electrons during the complete reaction.
Only \(8 mL\) of \(KMnO _{4}\) solution would be required to neutralise a \(20 mL\) equimolar solution of metal sulphite.

Hence option C is the correct answer.
Question 3
1 / -0

A non-volatile solute P dimerises in a solvent as 2P ⇌ P². If 'm' is the molality of the solute in the solvent, then the equilibrium constant of the given reaction is equal to :

A

\(\frac{K_{b}\left(K_{b} m+\Delta T_{b}\right)}{\left(2 \Delta T_{b}-K_{b} m\right)^{2}}\)

B

\(\frac{K_{b}\left(K_{b} m-\Delta T_{b}\right)}{\left(2 \Delta T_{b}-K_{b} m\right)^{2}}\)

C

\(\frac{K_{b}\left(K_{b} m-\Delta T_{b}\right)}{\left(2 \Delta T_{b}+K_{b} m\right)^{2}}\)

D

\(\frac{K_{b}\left(K_{b} m+\Delta T_{b}\right)}{\left(2 \Delta T_{b}+K_{b} m\right)^{2}}\)

Solution

\(\quad 2 A \rightleftharpoons A_{2}\)

\((1-a) \quad(a / 2) ; \quad\) Total amount \(=(1-a / 2)\)

The van't Hoff factor is \(i=\frac{\text { Total amount in solution }}{\text { Amount of } A \text { to start with }}=\frac{(1-\alpha / 2)}{1}=1-\alpha / 2\) or \(\alpha=2(1-i)\) Hence \(\quad K_{\mathrm{oq}}=\frac{m(\alpha / 2)}{m^{2}(1-\alpha)^{2}}=\frac{(1-i)}{m(2 i-1)^{2}}\) where \(i=\frac{\Delta \bar{T}_{b}}{K_{b} m}\)

\(K_{e q}=\frac{c \alpha^{2}}{(1-\alpha)}=\frac{K_{b}\left(K_{b} m-\Delta T_{b}\right)}{\left(2 \Delta T_{b}-K_{b} m\right)^{2}}\)

Hence option B is the correct answer.
Question 4
1 / -0

An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO₃ and the total volume was made to be 100 mL. A silver electrode was dipped in the solution and the emf of the cell set up Pt(s), H₂(g) | H⁺(1M) || Ag⁺ (aq) | Ag(s) was 0.62 V. If E^ocell = 0.80 V, then what is the percentage of Ag in the alloy?

[At 25^oC, RT/F = 0.06]

A
25

B
2.50

C
10

D
50

Solution

\(P t(s), H_{2}(g)\left|H^{+}(1 M) \| A g^{+}(a q)\right| A g(s)\)

\(H_{2} \rightarrow 2 H^{+}+2 e^{-} \quad\) (at anode)

\(2 A g^{+}+2 e^{-} \rightarrow 2 \mathrm{Ag}\) (at cathode)

\(2 A g^{+}+2 e^{-} \rightarrow 2 \mathrm{Ag}\) (at cathode)

\(H_{2}+2 A g^{+} \rightarrow 2 A g+2 H^{+}\)

(overall reaction) \(E_{c c l}=E^{\circ}-\frac{2.303 R T}{2 F} \log \frac{\left|H^{+}\right|^{2}}{\left|A g^{+}\right|^{2}\left|H_{2}\right|}\)
\(E_{\text {cell }}=E^{\circ}-\frac{2.303 R T}{2 F} \log \frac{1}{\left[A g^{+}\right]^{2}}\)
\(0.62=0.80-\frac{2.303 \times 0.06}{2} \log \frac{1}{\left|A g^{+}\right|^{2}}\)
\(0.62=0.80+\frac{2 \times 2.303 \times 0.06}{2} \log \left[A g^{+}\right]\)
\(-0.18=0.1382 \log \left[A g^{+}\right]\)
\(\left[A g^{+}\right]=0.05 m\)

\(\therefore\) Mole of \(A g^{+}\) in \(1000 \mathrm{mL}=0.05 \times \frac{100}{1000}\)

Wt. of \(A g^{+}\) in \(1000 m L=0.05 \times \frac{100}{1000} \times 108\)

\(\%\) of \(A g\) in \(1.08 g\) alloy

\(=\frac{0.05 \times 100 \times 108}{1000 \times 1.08} \times 100\)

\(=50 \%\)

Hence option D is the correct answer.
Question 5
1 / -0

Directions: The following question has four choices, out of which only one is correct.
Consider the cell Pt \(\left| H _{2}(0.75 atm )\right| HCl (0.25 M ) \| sn ^{4+}(0.60 M ), Sn ^{2+}(1.50 M ) \mid Pt\) Given \(E_{S n^{4}+/ S n^{2+}}^{\circ}=0.13 V ,\) which of the following statements is correct?

A
The cell reaction of the above cell is Sn²⁺ + 2H+→ Sn⁴⁺ + H₂.

B
E.M.F. of the cell is 2.15 V.

C
When Sn²⁺ | Sn⁴⁺ becomes 3.06 x 10⁵, the emf of the cell becomes zero.

D
The emf of cell is independent of the pH of the acid used.

Solution

\(E=0=0.13-(0.059) / 2\left[\log \left\{\frac{[.25]^{2}}{(.75)}\right\} \frac{\left[S n^{2+}\right]}{\left[S n^{4+}\right]}\right.\)
So, \(\frac{\left[S n^{2+}\right]}{\left[S n^{4+}\right]}=3.06 \times 10^{5}\)
Thus, this statement is correct.

Hence option C is the correct answer.
Question 6
1 / -0

Which of the following statements regarding the study of oxides of nitrogen is/are correct?

(i) Except N₂O₅, all oxides of nitrogen are endothermic as a large amount of energy is required to dissociate the stable molecules of oxygen and nitrogen.

(ii) Due to small electronegativity difference, N-O bond can be easily broken. Therefore, oxides of nitrogen are good reducing agents.

(iii) N₂O₃ is stable at only higher temperature.

(iv) All oxides of nitrogen are acidic in nature, except N₂O and NO.

A
Only (i)

B
(i) and (iv)

C
(i) and (ii)

D
All of these

Solution

(ii) Due to small electronegativity difference, N-O bond is easily breakable. Therefore, oxides of nitrogen are said to be better oxidising agents. Hence, (ii) is wrong.

(iii) N₂O₃ is stable at only lower temperature. Hence, (iii) is wrong.

N₂O₅ is a stable oxide of nitrogen.

N₂O and NO are neutral oxides.

Hence option B is the correct answer.
Question 7
1 / -0

What is the work done against the atmosphere when 25 grams of water vaprorizes at \(373 K\) against a constant external pressure of 1atm ? Assume that steam obeys perfect gas law. Given that the molar enthalpy of vaporization is \(9.72 \mathrm{kcal} / \mathrm{mole},\) what is the change of internal energy in the above process?

A
-1036.194 cal and 12,464.886 cal

B
921.4cal,11074cal

C
1025.6cal,12474.3cal

D
1129.3cal,10207cal

Solution

Mole of \(H_{2} O=1.39\)

\(P v=n R T\)

\(1 \times v=1.39 \times 0.082 \times 373\)

\(v=42.5\) llit

\(w=\) Pext. \(d v=1 \times[42.80]\) atm \(\times\) lit. \(=42.80 \times 101.325 J=\frac{42.80 \times 101.325}{4.2}=1025.6 \mathrm{cal}\)

\(\Delta H=\Delta U+[P \Delta v] .=12470.6 \mathrm{cal}\)

\(\Delta U=\Delta H-P \Delta v=13500-1025.6=1247.3 \mathrm{cal}\)

Hence option C is the correct answer.
Question 8
1 / -0

Directions: The following question has four choices, out of which only one is correct.

An organic compound containing 60.84% C, 40.35% H and the rest of O gives effervescence with NaHCO₃. It also gives deep violet colouration with FeCl₃ solution. The compound can be obtained from phenol by the action of CCl₄/KOH.

The compound contains

A
a phenolic OH group only

B
a carboxyl group only

C
both phenolic OH and carboxyl groups

D
None of these

Solution

This is correct as it gives tests for both carboxylic acid as well as phenolic groups and the compound is salicylic acid.

It can be prepared from the Riemann Tiemann reaction as follows:

Moreover, the molecular formula C₇H₆O₃ justifies the percentage of carbon, hydrogen and oxygen in the data. Hence option C is correct.
Question 9
1 / -0

The standard redox potentials \(\left( E ^{\circ}\right)\) of the following reactions are as follows.
I. \(MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O ; E ^{\circ}=1.51 V\)
II. \(Sn ^{2+} \rightarrow Sn ^{4+}+2 e ^{-} ; E ^{\circ}=-0.15 V\)
III. \(Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O ; E ^{\circ}=1.33 V\)
\(IV . Ce ^{3+} \rightarrow Ce ^{4+}+ e ^{-} ; \quad E ^{ O }=-1.61 V\)
In which of the following options is the oxidising power of the various ions wrongly related?

A
Cr₂O₇^2- > MnO^-4

B
Ce⁴⁺ > Sn⁴⁺

C
Ce⁴⁺ > MnO^-4

D
MnO^-4 > Sn⁴⁺

Solution

The higher the reduction potential, the higher the tendency to be reduced and higher the oxidising power. Correct order is \(MnO _{4}^{-}>Cr O _{7}^{-2}\)

Hence option A is the correct answer.
Question 10
1 / -0

How many litres of hydrogen gas can be generated by reacting 6.25 g of barium hydride with water at 20^oC and 755 mm Hg pressure, according to the following chemical equation?

BaH₂(s) + 2H₂O(l) → Ba(OH)₂(aq) + 2H₂(g)

A
1.60 L

B
2.158 L

C
0.799 L

D
0.2158 L

Solution

\begin{equation}\begin{aligned}
&BaH _{2}( s )+2 H _{2} O ( I ) \rightarrow Ba ( OH )_{2}( aq )+2 H _{2}( g )\\
&n_{B a H_{2}}=\frac{w}{M}=\frac{6.25}{140}=0.0446\\
&n_{H_{2}}=0.0446 \times 2=0.0892\\
&\text { Volume of the } H _{2} \text { formed at } STP =0.0892 \times 22.4=1.998 L\\
&\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\\
&\frac{755 \times V_{2}}{293}=\frac{760 \times 1.998}{273}\\
&V_{2}=2.158 L
\end{aligned}\end{equation}

Hence option B is the correct answer.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test - 8

Result

Chemistry Test - 8

Score

-

Rank

-

SHARING IS CARING

Question 1

The acidity of hydroxyl amine is 1.

Equivalent weight of Fe2(SO4)3 is M/2.

6 milliequivalent of Fe2(SO4)3 is contained in 3 millimoles of ferric sulphate.

All of the above

Solution

Question 2

1 mole of potassium permanganate oxidises 5 moles of the metal sulphite.

There is a net transfer of 5 electrons in the reaction.

20 mL of 0.5M KMnO4 solution can neutralise 10 mL of 0.25 M metal sulphite.

50 mL of KMnO4 can oxidise a 20 mL equimolar solution of the metal sulphite.

Solution

Question 3

\(\frac{K_{b}\left(K_{b} m+\Delta T_{b}\right)}{\left(2 \Delta T_{b}-K_{b} m\right)^{2}}\)

\(\frac{K_{b}\left(K_{b} m-\Delta T_{b}\right)}{\left(2 \Delta T_{b}-K_{b} m\right)^{2}}\)

\(\frac{K_{b}\left(K_{b} m-\Delta T_{b}\right)}{\left(2 \Delta T_{b}+K_{b} m\right)^{2}}\)

\(\frac{K_{b}\left(K_{b} m+\Delta T_{b}\right)}{\left(2 \Delta T_{b}+K_{b} m\right)^{2}}\)

Solution

Question 4

25

2.50

10

50

Solution

Question 5

The cell reaction of the above cell is Sn2+ + 2H+→ Sn4+ + H2.

E.M.F. of the cell is 2.15 V.

When Sn2+ | Sn4+ becomes 3.06 x 105, the emf of the cell becomes zero.

The emf of cell is independent of the pH of the acid used.

Solution

Question 6

Only (i)

(i) and (iv)

(i) and (ii)

All of these

Solution

Question 7

-1036.194 cal and 12,464.886 cal

921.4cal,11074cal

1025.6cal,12474.3cal

1129.3cal,10207cal

Solution

Question 8

a phenolic OH group only

a carboxyl group only

both phenolic OH and carboxyl groups

None of these

Solution

Question 9

Cr2O72- > MnO-4

Ce4+ > Sn4+

Ce4+ > MnO-4

MnO-4 > Sn4+

Solution

Question 10

1.60 L

2.158 L

0.799 L

0.2158 L

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

Equivalent weight of Fe₂(SO₄)₃ is M/2.

6 milliequivalent of Fe₂(SO₄)₃ is contained in 3 millimoles of ferric sulphate.

20 mL of 0.5M KMnO₄ solution can neutralise 10 mL of 0.25 M metal sulphite.

50 mL of KMnO₄ can oxidise a 20 mL equimolar solution of the metal sulphite.

The cell reaction of the above cell is Sn²⁺ + 2H+→ Sn⁴⁺ + H₂.

When Sn²⁺ | Sn⁴⁺ becomes 3.06 x 10⁵, the emf of the cell becomes zero.

Cr₂O₇^2- > MnO^-4

Ce⁴⁺ > Sn⁴⁺

Ce⁴⁺ > MnO^-4

MnO^-4 > Sn⁴⁺