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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three

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2 months ago

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

In alanylglycylleucylalanylvaline, the number of peptide linkages is____________.

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Payal Gupta

Contributor-Level 10

Ala – Gly – Leu-Ala- Val

Two amino acid units are connected by peptide linkage. So number of peptide linkage = 4.

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

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Payal Gupta

Contributor-Level 10

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

0.2g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N₂ evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is __________. [nearest integer]

(Given : Molar mass of N₂ is 28g mol^-1, Molar volume of N₂ at STP : 22.4L)

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Payal Gupta

Contributor-Level 10

Number of moles of gas = $\frac{22.4}{22400} = \frac{1}{1000} m o l e s$

Mass of nitrogen = $(\frac{1}{1000} * 28) g m$

% of N in organic compound = $\frac{(\frac{28}{1000})}{0.2} * 100 % = 14 %$

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three 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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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three 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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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three 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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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three Chemistry Electrochemistry

The resistance of a conductivity cell containing 0.01 MKCl solution at 298K is $1750 Ω$ . If is 0.152 * 10^-3 cm^-1, then the cell constant of the conductivity cell _________ * 10^-3 cm^-1.

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Payal Gupta

Contributor-Level 10

$K = \frac{1}{R} * c e l l c o n s t a n t$

= 0.152 * 10^-3 * 1750

= 266 * 10^-3 cm^-1

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three Chemical Equilibrium

PCl₅ dissociates as

PCl₅(g) $?$ PCl₃ (g) + Cl₂(g)

5 moles of PCl₅ placed in a 200 litre vessel which contains 2 moles of N₂ and is maintained at 600K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl₅ is ____________ * 10^-3g. (nearest integer).

(Given R = 0.082 L atm K^-1 mol^-1; Assume ideal gas behavior)

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Payal Gupta

Contributor-Level 10

$\begin{array}{l} P C l_{5} (g) ? P C l_{3} (g) + C l_{2} (g) \\ t = 0 5 m o l e s 0 0 \\ \underline{- x m o l e} \underline{+ x m o l e} + \underline{x m o l e} \end{array}$

Eq.(5 – x) mole x molex mole

Total number of moles at equilibrium

$= (5 - x) + x + x + n_{N_{2}} = (7 + x)$

$K_{p} = \frac{(P_{P C l_{3}}) * {(P C l_{2})}_{C l}}{{(P_{P C l_{5}})}_{e q}} = \frac{(\frac{3}{10} * 2.46) * (\frac{3}{10} * 2.46)}{(\frac{2}{10} * 2.46)} = 1.107 a t m$

= 1107 * 103 atm

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three Chemistry Solutions

A Company dissolves 'x' amount of CO₂ at 298K in 1 litre of water to prepare soda water X = _________________ * 10^-3 g. (nearest integer)

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Payal Gupta

Contributor-Level 10

According to Henry's law

$P_{G a s} = K_{H} * x_{G a s}$

$= 0.835 = 1.67 * 1 0^{3} * [\frac{n_{C O_{2}}}{n_{C O_{2}} + 55.5}]$

As $n_{C O_{2}} < < 55.5$

$\Rightarrow n_{C O_{2}} = 27.78 * 1 0^{- 3} m o l e s$

Mass of CO₂ gas = $n_{C O_{2}} * (M M) C O_{2}$

= 1222 * 10^-3gm

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

At 300K, a sample of 3.0g of gas A occupies the same volume as 0.2g of hydrogen at 200K at the same pressure. The molar mass of gas A is _________g mol^-1, (nearest integer) Assume that the behavior of gases as ideal. (Given: The molar mass of hydrogen (H₂) gas is 2.0g mol^-1.)

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Payal Gupta

Contributor-Level 10

${P V)}_{H_{2}} = {(P V)}_{G a s}$

${(n R T)}_{H_{2}} = {(n R T)}_{G a s}$

$n_{H_{2}} * T_{H_{2}} = n_{G a s} * T_{G a s}$

$\Rightarrow \frac{0.2}{2} * 200 = \frac{3}{{(M M)}_{G a s}} * 300 \Rightarrow {(M M)}_{G a s} = 45 g m / m o l e$

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