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11th Standard Syllabus & Materials

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model test Set B

Published on: 02/06/2019

Question Set B

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Download Tamil Nadu 11th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

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Std: 11th Standard - Chemistry

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Questions:

Multiple Choice Question

6 x 1 = 6

The equivalent mass of potassium permanganate in alkaline medium is:
MnO₄^- + 2H₂O + 3e^-\(\rightarrow\) MnO₂ + 4OH^-

(a)

31.6

(b)

52.7

(c)

(d)

None of these

If Avogadro number were changed from 6.022 x 10²³to 6.022 x 10²⁰, this would change ______.

(a)

the ratio of chemical species to each other in a balanced equation

(b)

the ratio of elements to each other in a compound

(c)

the definition of mass in units of grams

(d)

the mass of one mole of carbon

The oxidation number of Cr in Cr₂O₇^2- _______ is

(a)

(b)

-6

(c)

(d)

-7

The oxidation number of hydrogen in LiH is _________

(a)

(b)

-1

(c)

(d)

-2

Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidising behaviour ?

(a)

Cu + 2H₂SO₄ \(\longrightarrow \) CuSO₄ +SO₂ + 2H₂O

(b)

C + 2H₂ + SO₄ \(\longrightarrow \) CO₂ + 2SO₂ + 2H₂O

(c)

BaCl₂ + H₂SO₄ \(\longrightarrow \) BaSO₄ + 2HCl

(d)

None of the above

An element X has the following isotopic Composition ²⁰⁰X = 90%, ¹⁹⁹X = 8% and ²⁰²X=2%.The Weighted average atomic mass of the element X is closest to _________.

(a)

201 u

(b)

202 u

(c)

199 u

(d)

200 u

Generic 3

1 x 3 = 3

Which contains the greatest number of moles of oxygen atoms
i) 1 mol of ethanol
ii) 1 mol of formic acid
iii) 1 mol of H₂O

Generic 5

3 x 5 = 15

Explain the term competitive electron transfer reaction with an example.

Balance the following equations by oxidation number method
i) \({ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+KI+{ H }_{ 2 }SO_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }({ SO }_{ 4 })+{ I }_{ 2 }+{ H }_{ 2 }O\)
ii) \({ K }Mno_{ 4 }+{ Na }_{ 2 }{ So }_{ 3 }\longrightarrow { MnO }_{ 2 }+{ Na }_{ 2 }{ So }_{ 4 }+KOH\)
iii) \(Cu+{ HNO }_{ 3 }\longrightarrow Cu\left( { No }_{ 3 } \right) _{ 2 }+{ No }_{ 2 }+{ H }_{ 2 }O\)
iv) \({ KMn }O_{ 4 }+{ H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ MnSO }_{ 4 }+{ CO }_{ 2 }+{ H }_{ 2 }O\)

10.

A Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and 0 = 69.5%.
Calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as a water of crystallization. (molecular mass of the compound is 322).

Generic 2

1 x 2 = 2

11.

Balance the following equations by oxidation number method - \({ K }Mno_{ 4 }+{ Na }_{ 2 }{ So }_{ 3 }\longrightarrow { MnO }_{ 2 }+{ Na }_{ 2 }{ So }_{ 4 }+KOH\)

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Answers Key:

Multiple Choice Question

6 x 1 = 6

(b)
52.7

(d)
the mass of one mole of carbon

(a)
+6

(b)
-1

(d)
200 u

Generic 3

1 x 3 = 3

Compound	Given No.of moles	No.of oxygen atoms
Ethanol - C₂H₅OH	1	1\(\times\)6.022\(\times\)10²³
Formic acid - HCOOH	1	2\(\times\)6.022\(\times\)10²³
Water - H₂O	1	1\(\times\)6.022\(\times\)10²³
Formic acid

Generic 5

3 x 5 = 15

When a strip of metallic copper in sliver nitrate solution ~aken in a beaker and after some time, the solution slowly turns blue. This is due to the formation of Cu²⁺ ions, i.e. copper replaces silver from silver nitrate. The reaction is,

It indicates that between copper and silver, copper has the tendency to release electrons and silver to accept electrons. This type of metal displacement reactions are known as competitive electron transfer reactions.

(i) \({ K }_{ 2 }\overset { +6 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \uparrow } }{ Cr_{ 2 } } } { O }_{ 7 }+K\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \downarrow }{ I } } } +{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ \overset { +3 }{ Cr } }_{ 2 }({ SO }_{ 4 })_{ 3 }+\overset { 0 }{ I } _{ 2 }+{ H }_{ 2 }O\)
K₂Cr₂O₇ + 6KI + H₂SO₄ \(\longrightarrow \) K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
K₂Cr₂O₇ + 6KI + H₂SO₄ \(\longrightarrow \) K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
K₂Cr₂O₇ + 6KI + 7H₂SO₄ \(\longrightarrow \) 4k₂SO₄ + Cr₂(SO₄)₃ + ₃I₂ + 7H₂
ii) \({ K }Mno_{ 4 }+{ Na }_{ 2 }{ So }_{ 3 }\longrightarrow { MnO }_{ 2 }+{ Na }_{ 2 }{ So }_{ 4 }+KOH\)
\({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \) MnO₂ + Na₂SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \) 2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂SO₄ + 2KOH
iii) \(Cu+{ HNO }_{ 3 }\longrightarrow Cu\left( { No }_{ 3 } \right) _{ 2 }+{ No }_{ 2 }+{ H }_{ 2 }O\)
\(\overset { 0 }{ \underset { \underset { 2e^{ - } }{ \downarrow } }{ Cu } } { O }_{ 7 }+H\overset { +5 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ N } } } { O }_{ 3 }\longrightarrow \overset { +2 }{ Cu } \left( { No }_{ 3 } \right) _{ 2 }+\overset { +4 }{ N } { O }_{ 2 }+{ H }_{ 2 }O\)
Cu +2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ \(\longrightarrow \) Cu (NO₃)₂ + 2No₂ + 2H₂O
iv) \({ KMn }O_{ 4 }+{ H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ MnSO }_{ 4 }+{ CO }_{ 2 }+{ H }_{ 2 }O\)
\({ K }\overset { +7 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \downarrow } }{ M } } n{ O }_{ 4 }+{ H }_{ 2 }\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ C_{ 2 } } } } { O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+\overset { +2 }{ M } n{ SO }_{ 4 }+\overset { +4 }{ C } { O }_{ 2 }+{ H }_{ 2 }O\)
2KMnO₄ + 5 H₂C2O₄ + H₂S0₄ \(\longrightarrow \) Mn0₂ + Na₂S0₄ + KOH
2KMnO₄+ 5 H₂C₂O₄ + H₂S0₄ \(\longrightarrow \) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O
2KMnO₄ + 5 H₂C₂O₄ + 3H₂S0_{4 \(\longrightarrow \)} K₂S0₄ + 2MnS0₄ + 10C0₂ + 8 H₂0

10.

Element	%	Relative number of atoms	Simple Ratio
Na	14.31	\(\frac { 14.31 }{ 23 } =0.62\)	\(\frac { 0.62 }{ 0.31 } =2\)
S	9.97	\(\frac { 9.97 }{ 32 } =0.31\)	\(\frac { 0.31 }{ 0.31 } =1\)
H	6.22	\(\frac { 6.22 }{ 1 } =6.22\)	\(\frac { 6.22 }{ 0.31 } =20\)
O	69.5	\(\frac { 69.5 }{ 16 } =4.34\)	\(\frac { 4.34 }{ 0.31 } =14\)

Empirical formula = Na₂SH₂₀O₁₄
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄10H₂O.

Generic 2

1 x 2 = 2

11.

(ii) \({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \) MnO₂ + Na₂SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \) 2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂SO₄ +2KOH