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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 11th

The correct order of density is

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Payal Gupta

Contributor-Level 10

Element Density (gm/cm³)

Be 1.84

Mg 1.74

Ca &nb

...more

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 11th

High purity (> 99.95%) dihydrogen is obtained by

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Payal Gupta

Contributor-Level 10

High purity (> 99.95%) H₂ gas is obtained by electrolysis of warm aqueous Ba (OH)₂ solution using Ni electrode.

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 12th

Given below are two statements.

Statement I: Pig iron is obtained by heating cast iron with scrap iron.

Statement II: Pig iron has a relatively lower carbon content than that of cast iron. In the light of the above statements, choose the correct answer form the options given below.

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Payal Gupta

Contributor-Level 10

Cast iron is formed by pig iron and scrap iron.

Pig Iron → Carbon 4%

Cast Iron → Carbon 3%

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 11th

The first ionization entahalpies of Be, B, N and O follow the order

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Payal Gupta

Contributor-Level 10

Ionization enthalpy: B < Be < O < N

IE of N > O [Due to half-filled configuration]

IE of Be > B [Due to penetration effect]

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 12th

For micelle formation, which of the following statements are correct?

A. Micelle formation is an exothermic process.

B. Micelle formation is an endothermic process.

C. The entropy change is positive.

D. The entropy change is negative.

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Payal Gupta

Contributor-Level 10

During micelle formation, $Δ S$ is positive

Micelle formation is endothermic at low temperature and exothermic at high temperature.

At room temperature ⇒ Micelle formation is endothermic.

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 12th

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is

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Payal Gupta

Contributor-Level 10

$\land_{M} = \frac{K * 1000}{M}$ $M_{2} = \frac{20}{(80 / 1000)}$

$\frac{\land_{M_{1}}}{\land_{M_{2}}} = \frac{M_{2}}{M_{1}}$ $M_{1} = \frac{10}{(20 / 1000)}$

$\frac{\land_{M_{1}}}{\land_{M_{2}}} = \frac{1}{2} \Rightarrow \land_{M_{2}} = 2 \land_{M_{1}}$

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 11th

$K_{a_{1}}, K_{a_{2}} a n d K_{a_{3}}$ are the respective ionization constants for the following reaction (a), (b) and (c).

(A) $H_{2} C_{2} O_{4} ? H^{+} + H C_{2} O_{4}^{-}$

(B) $H C_{2} O_{4}^{+} ? H^{-} + C_{2} O_{4}^{2 -}$

(C) $H_{2} C_{2} O_{4} ? 2 H^{+} + C_{2} d_{4}^{2 -}$

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Payal Gupta

Contributor-Level 10

(I) $H_{2} C_{2} O_{4} ? H^{+} + H C_{2} O_{4}^{-}$

(II) $H C_{2} O_{4}^{-} ? H^{+} + C_{2} O_{4}^{2 -}$

(I) + (II); $H_{2} C_{2} O_{4} ? 2 H^{+} + C_{2} O_{4}^{2 -}$

$K_{a_{3}} = K_{a_{1}} * K_{a_{2}}$

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Chemistry Solutions

Two solutions A and B are prepared by dissolving 1g of no-volatile solutes X and Y respectively in 1 kg of water. The ratio of depression in freezing point for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is

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Payal Gupta

Contributor-Level 10

$Δ T_{f} = i * k_{f} * M$

$\frac{{(Δ T_{f})}_{x}}{{(Δ T_{f})}_{y}} = \frac{i K_{f} m_{x}}{i K_{f} M_{Y}} = \frac{M_{X}}{M_{Y}}$

$\Rightarrow \frac{1}{4} = \frac{1}{M_{X}} * \frac{M_{Y}}{1} \Rightarrow \frac{M_{X}}{M_{Y}} = \frac{4}{1} = \frac{1}{0.25}$

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 12th

Match List I with List II:

List I (molecule)		List II (hybridization; shape)
A.	XeO₃	I.	sp³d; linear
B.	XeF₂	II.	sp³; pyramidal
C.	XeOF₄	III.	sp³d³; distorted octahedral
D.	XeF₆	IV.	sp³d³; square pyramidal

Choose the correct answer from the options given below

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Payal Gupta

Contributor-Level 10

$X e O_{3} \Rightarrow s p^{3}, P y r a m i d a l$

XeF₂ sp³d₂, square pyramidal

XeF₆sp³d³, Distorted octahedral

XeOF₄ sp³d², square pyramidal

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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter Two Class 12th

The moles of methane required to produce 81 g of water after complete combustion is________* 10^-2 mol. (nearest integer)

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Vishal Baghel

Contributor-Level 10

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

2 moles of water produced by 1 mole of methane

Or 36 gm of water produced by 1 mole of CH₄

$∴$ 81 gm of water produced = $\frac{1 * 81}{36}$ = 2.25 mole of CH₄

Mole of CH4 required = 225 * 10^-2

the nearest integer = 225

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