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Ncert Solutions Chemistry Class 12th

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Ncert Solutions Chemistry Class 12th Class 12th

For a given chemical reaction

$γ_{1} A + γ_{2} B \to γ_{3} C + γ_{4} D$

Concentration of C charges form 10 mmol dm^-³ to 20 mmol dm^-³ in 10 second. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm^-³ s^-¹. Therefore the rate of reaction is_____________mmol dm^-³ s^-¹ (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$\frac{d [C]}{d t} = (\frac{20 - 10}{10}) = 1 m m o l d m^{- 3}$

$+ \frac{d [D]}{d t} = {- \frac{d (B)}{d t}} * 1.5 = \frac{3}{2} {- \frac{d [B]}{d t}}$

${- \frac{d [B]}{d t}} = 2 * {\frac{- d [A]}{d t}}$

$∴ \frac{1}{6} {\frac{- d [B]}{d t}} = \frac{1}{3} {\frac{- d [A]}{d t}} = \frac{1}{9} {\frac{+ d [D]}{d t}} = \frac{d [C]}{d t}$

$∴$ rate of reaction = $\frac{+ d [C]}{d t}$ = 1m.m dm^-3 S^-1

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Ncert Solutions Chemistry Class 12th Class 12th

The distance between Na⁺ and Cl^- ions in solid NaCl of density 43.1g cm^-³ is ________×10^-¹⁰m. (Nearest Integer)

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Ncert Solutions Chemistry Class 12th Class 12th

Among the following, which is the strongest oxidizing agent?

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Payal Gupta

Contributor-Level 10

$E_{M n^{+ 3} / M n^{+ 2}}^{o} = 1.51 V$

the strongest oxidizing agent have the highest reduction potential. So Mn³⁺ is the strongest oxidizing agent.

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Ncert Solutions Chemistry Class 12th Class 12th

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?

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Payal Gupta

Contributor-Level 10

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l} C e^{+ 4} + e^{-} \to C e^{+ 3} E^{0} = 1.61 V \\ C e^{+ 3} + 3 e^{-} \to C e E^{0} = - 2.336 V \end{array}$

It exist as Ce+4 and acts like a strong oxidizing agent by gaining electrons

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Ncert Solutions Chemistry Class 12th Class 12th

Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold an another complex [B], [A] and [B], respectively are

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Payal Gupta

Contributor-Level 10

Au + NaCN + O₂ → Na [Au (CN)₂]

$Z n + N a [A u {(C N)}_{2}] \to N a_{2} [Z n {(C N)}_{4}] + A u$

$A i s {[A u {(C N)}_{2}]}^{-} a n d B i s {[Z n {(C N)}_{4}]}^{- 2}$

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Ncert Solutions Chemistry Class 12th Class 12th

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Ncert Solutions Chemistry Class 12th Class 12th

Incorrect statement for Tyndall effect is:

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Payal Gupta

Contributor-Level 10

$A l^{+ 3}$ → number of electrons = 10

Br → number of electrons = 36

$B e^{+ 2}$ → number of electrons = 2

$C l^{-}$ → number of electrons = 18

$L i^{+}$ → number of electrons = 2

$S^{- 2}$ → number of electrons = 18

K⁺ → number of electrons = 18

O^-2 → number of electrons = 10

Mg⁺² → number of electrons = 10

O^-2 & Mg⁺² are isoelectronic with Al⁺³

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Ncert Solutions Chemistry Class 12th Class 12th

Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is ____________.

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Payal Gupta

Contributor-Level 10

In acidic solution

$3 \overset{+ 6}{M n} O_{4}^{2 -} + 4 H^{+} \to 2 \overset{+ 7}{M n} O_{4}^{-} + \overset{+ 4}{M n} O_{2} + 2 H_{2} O$

Difference in oxidation state of Mn is product

= 7 – 4 = 3

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Ncert Solutions Chemistry Class 12th Class 12th

(a) Baryte (b) Galena (c) Zinc blende and (d) Copper pyrites. How many of these minerals are sulphide based?

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Payal Gupta

Contributor-Level 10

Baryte – BaSO₄ [Sulphate based ore]

Galena – PbS

Zinc blende – ZnS

Copper pyrite – CuFeS₂

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Ncert Solutions Chemistry Class 12th Class 12th

When 200mL of 0.2 M acetic acid is shaken with 0.6g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1M. The mass of acetic acid adsorbed per gram of carbon is ______________ g.

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Payal Gupta

Contributor-Level 10

Number of moles of CH₃COOH Adsorbed on 0.6 gm wood charcoal.

$= (0.2 * 200) * 1 0^{- 3} - (0.1 * 200) * 1 0^{- 3}$

= 20 * 10^-3 moles

Mass of CH₃COOH absorbed = 20 * 10^-3 * 60 = 1.2 gm

Mass of CH₃COOH adsorbed per gram = $\frac{1.2}{0.6} = 2 g$ of charcoal.

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