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Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

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

2.0g of H₂ gas is adsorbed on 2.5g platinum powder at 300K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______________mL.

(Given : R = 0.083 L bar K^-¹ mol^-¹)

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Payal Gupta

Contributor-Level 10

Volume of H₂adsorbed = $\frac{n R T}{P} = \frac{2 * 0.083 * 300}{2 * 1} = 24.9 l i t = 24900 m l$

Therefore volume of gas adsorbed per gram of the adsorbent = $\frac{24900}{2.5} = 9960$

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

A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s.

(Given : In 2 = 0.693)

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Payal Gupta

Contributor-Level 10

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100} * t = 2.303 l o g_{10} \frac{A_{0}}{A_{t}}$ ………….(i)

and $\frac{0.693}{50} * t = 2.303 l o g_{10} \frac{B_{0}}{B_{t}}$ ………….(ii)

from equation (i) and (ii)

$0.693 t [\frac{1}{50} - \frac{1}{100}] = [l o g \frac{B_{0}}{B_{t}} - l o g \frac{A_{0}}{A_{t}}] * 2.303$

Here; A0 = B0 and $A_{t} = 4 * B_{t}$

Therefore $0.693 t [\frac{1}{100}] = 2.303 [l o g (\frac{B_{0}}{B_{t}} * \frac{A_{t}}{A_{0}})]$

$∴ t = \frac{2.303 * 0.3010 * 2 * 100}{693} = 200 s$

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

50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10^-². (Nearest integer)

(Given : pKa (CH₃-COOH) = 4.76)

log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04

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Payal Gupta

Contributor-Level 10

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$p H = p K a + l o g_{10} \frac{[S]}{[A]}$

$p H = 4.76 + l o g_{10} \frac{2.5}{2.5} = 4.76 = 476 * 1 0^{- 2}$

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Six Structure of Atom

A 0.5 percent solution of potassium choride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is__________. (Nearest integer)

(Molal depression constant for water is 1.80 K kg mol^-¹ and molar mass of KCl is 74.6 g mol^-¹)

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Payal Gupta

Contributor-Level 10

0.5 % KCl solution has molality (m) = $\frac{0.5 * 1000}{74.5 * 99.5}$

$K C l_{(a q)} ? K_{(a g)}^{+} + C l_{(a q)}^{-}$

1 - α α α

And I = $(1 - α + α + α) = 1 + α$

$i = \frac{Δ T f}{k f * m} = \frac{0.24 * 74.5 * 99.5}{1.8 * 0.5 * 1000} = 1 + α$

1.976 = 1 + α

$∴ α = 0.976$

% = 97.6%

the nearest 98.

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Six Chemistry Some Basic Concepts of Chemistry

For complete combustion of methanol

$C H_{3} O H (1) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (I)$

The amount of heat produced as measured by bomb calorimeter is 726 kJmol^-¹ at 27°C. The enthalpy of combustion for the reaction is -x kJ mol^-¹, where x is_________

(Nearest integer)

(Given : R = 8.3 J K^-¹ mol^-¹).

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