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Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

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3 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Chemistry Chemical Kinetics

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is

(Given : In 10 = 2.303 and log 2 = 0.3010)

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alok kumar singh

Contributor-Level 10

$T_{90 %} = \frac{2.303}{k} l o g \frac{100}{10}$

$T_{50 %} = \frac{2.303}{k} l o g \frac{100}{50}$

$\frac{T_{90 %}}{T_{50 %}} = \frac{l o g 10}{l o g 2} = \frac{1}{0.301} = 3.32$

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3 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Chemistry Chemical Bonding and Molecular Structure

The correct order of bond orders of is $C_{2}^{2 -}, N_{2}^{2 -} a n d O_{2}^{2 -}$ , respectively

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alok kumar singh

Contributor-Level 10

Bond order of $C_{2}^{2 -} = \frac{N_{B} - N_{A}}{2} = \frac{10 - 4}{2} = 3$

Bond order of $N_{2}^{2 -} = \frac{N_{B} - N_{A}}{2} = \frac{10 - 6}{2} = 2$

Bond order of $O_{2}^{2 -} = \frac{N_{B} - N_{A}}{2} = \frac{10 - 8}{2} = 1$

N_B = No. of electrons in bonding molecular orbitals.

N_A= No. of electron is Anti bonding molecular orbitals

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3 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Structure of Atom

The energy of one mole of photons of radiation of wavelength 300nm is (Given: h = 6.63 * 10³⁴ Js,N_A = 6.02 * 10²³ mol^-1 c = 3 * 10⁸ ms^-1)

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alok kumar singh

Contributor-Level 10

Energy of 1 photon = $\frac{h c}{λ}$

Energy of 1 mole of photon = $N_{A} * \frac{h c}{λ}$ $_{} \frac{}{}$

$= \frac{6.022 * 1 0^{23} * 6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{300 * 1 0^{- 9}}$

= 0.399 * 106 J = 399 KJ

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3 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Chemistry Some Basic Concepts of Chemistry

120g of an organic compound that contains only carbon and hydrogen gives 330g of CO₂ and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

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alok kumar singh

Contributor-Level 10

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 * no. of moles of H₂O = $(\frac{270}{18} * 2)$ moles

Mass of C = $\frac{330}{44} * 12$ gm = 90 gm

Mass of H = $\frac{270}{18} * 2 * 1 g m = 30 g m$

$\begin{array}{l} % o f C = \frac{90}{120} * 100 % = 75 % \\ % o f H = \frac{30}{120} * 100 % = 25 % \end{array}$

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3 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Electrochemistry

20mL of 0.02 M hypo solution is used for the titration of 10mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of Cu²⁺ is found to be__________* 10^-2 M. [nearest integer]

Given: 2Cu²⁺ + 4I^- ® Cu₂I₂ + I₂

I₂ + $2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}$

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alok kumar singh

Contributor-Level 10

$m e q o f l_{2} = m e q o f N a_{2} S_{2} O_{3} = 20 * 0.002 * 1$

2 * mmol of l₂ = 0.4

mmol of l₂ = 0.2 mmol

mmol of Cu²⁺ = 0.2 * 2 * 10^-3

$[C u^{2 +}] = \frac{0.4 * 1 0^{- 3}}{10 * 1 0^{- 3}} = 0.04 = 4 * 1 0^{- 2} M$

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4 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Class 11th

Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.

Column (I)	Column (II)
Electronic configuration	Electron gain enthalpy/kJ mol^–1
(i) 1s²2s²sp⁶	(A) – 53
(ii) 1s² 2s² 2p⁶ 3s¹	(B) – 328
(iii) 1s²2s² 2p⁵	(C) – 141
(iv) 1s² 2s² 2p⁴	(D) + 48

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

i d + 48

ii a &n

...more

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4 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Class 11th

Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.

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4 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Class 11th

Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.

Elements		?H₁	? H₂	?_egH
(i) Most reactive non metal	A	419	3051	– 48
(ii) Most reactive metal	B	i b	i b	i b
(iii) Least reactive element	C	738	1451	– 40
(iv) Metal forming binary halide	D	2372	5251	+ 48

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

i ) b 1681 3374 – 328

ii ) a 419 3051 – 48

iii)

...more

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4 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Class 11th

Match the correct atomic radius with the element.

Element	Atomic Radius (pm)
Be	74
C	88
O	111
B	77
N	66

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

Be 111

...more

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4 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Three Class 11th

Assertion (A) : Electron gain enthalpy becomes less negative as we go down a group.

Reason (R) : Size of the atom increases on going down the group and the added electron would be farther from the nucleus.

(i) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(ii) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(iii) Assertion and reason both are wrong statements.

(iv) Assertion is wrong statement but reason is correct statement.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (ii)

The electron gain enthalpy decreases down the group due to the increase in the atomic size which leads to decrease in the effective nuclear charge.

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