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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

0.01 M KMnO₄ solution was added to 20.0mL of 0.05 M Mohr's salt through a burette. The initial reading of 50mL burette is zero. The volume of KMnO₄ solution left in the burette after the end point is__________ mL. (nearest integer)

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alok kumar singh

Contributor-Level 10

n-factor of KMnO₄ in acidic medium = 5

n-factor of Mohr's salt = 1

meq of KMnO₄ = meq of Mohr's salt

0.01 * 5 * V = 0.05 * 1 * 20

Volume of KMnO₄ used, V = 20 mL

So; Volume of KMnO₄ left in burette = 50 -20mL

= 30 mL

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

The complete combustion of 0.492g of an organic compound containing 'C', 'H' and 'O' given 0.793g of CO₂ and 0.442g H₂O. The percentage of oxygen composition in the organic compound is __________. (nearest integer)

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alok kumar singh

Contributor-Level 10

Moles of C = moles of CO₂

= $\frac{0.793}{44} m o l$

Mass of C = $\frac{0.793}{44} * 12 g$

= 0.261g

Moles of H = 2 * $\frac{0.442}{18} m o l$

Mass of H = $2 * \frac{0.442}{18} * 1 g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492} * 100$

= 46.14%

the nearest integer = 46

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

The major product of the following reaction contains _____________ bromine atom(s).

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alok kumar singh

Contributor-Level 10

Bromination through free radical mechanism occurs at allylic carbon.

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

(a) CoCl₃.4NH₃, (b) CoCl₃.5NH₃, (c) CoCl₃. 6NH₃ and (d) CoCl(NO₃)₂. 5NH₃.

Number of complex(es) which will exist in cis-trans form is /are__________.

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alok kumar singh

Contributor-Level 10

(a) $C o C l_{3} . 4 N H_{3} i s [C o C l_{2} {(N H_{3})}_{4}] C l$ will show cis- trans isomerism as;

(b) $C o C l_{3} . 5 N H_{3} i s [C o {(N H_{3})}_{5} C l] C l_{2}$ will not show cis- trans isomerism.

(c) $C o C l_{3} . 6 N H_{3} i s [C o {(N H_{3})}_{6}] C l_{3}$ will not show cis- trans isomerism.

(d) $C o C l {(N O_{3})}_{2} . 5 N H_{3} i s [C o {(N H_{3})}_{6} C l] {(N O_{3})}_{2}$ will not show cis-trans isomerism.

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three Chemical Kinetics

${[F e {(C N)}_{6}]}^{4 -}$

${[F e {(C N)}_{6}]}^{3 -}$

${[T i {(C N)}_{6}]}^{3 -}$

${[N i {(C N)}_{4}]}^{2 -}$

${[C o {(C N)}_{6}]}^{3 -}$

Among the given complexes, number of paramagnetic complexes in__________.

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alok kumar singh

Contributor-Level 10

CN is a strong field ligand, so pairing occurs.

1. ${[F e {(C N)}_{6}]}^{4 -}$

So; it is diamagnetic

2. ${[F e {(C N)}_{6}]}^{3 -}$

So; it is paramagnetic.

3. ${[T i {(C N)}_{6}]}^{3 -}$

Ti³⁺ = 4s⁰3d¹

It is paramagnetic

4. ${[N i {(C N)}_{4}]}^{2 -}$

It is diamagnetic

5. ${[C o {(C n)}_{6}]}^{3 -}$

Hence, ${[F e {(C N)}_{6}]}^{3 -} a n d {[T i {(C N)}_{6}]}^{3 -}$ ${_{}}^{} {_{}}^{}$ are paramagnetic.

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

A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is___________. (Nearest integer)

(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)

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alok kumar singh

Contributor-Level 10

$t_{1 / 2} = 200 d a y s = \frac{0.693}{k}$

$t = \frac{2.303}{k} l o g_{10} \frac{N_{0}}{N}$

83 = $\frac{2.303}{0.693} * 200 l o g \frac{N_{0}}{N}$

$83 = \frac{200}{0.3010} l o g \frac{N_{0}}{N}$

0.125 = $l o g \frac{N_{0}}{N}$

$\frac{N_{0}}{N} = a n t i l o g (0.125)$

= 1.333

Activating remaining = $\frac{N}{N_{0}} * 100$

$= \frac{\frac{N_{0}}{1.333}}{N_{0}} * 100$

= 75%

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

2.5g protein containing only glycine (C₂H₅NO₂) is dissolved in water to make 500 mL of solution. TH osmotic pressure of this solution at 300 K is found to be 5.03 * 10^-3 bar. The total number of glycine units present in the protein is___________.

(Given : R = 0.083 L bar K^-1 mol^-1)

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alok kumar singh

Contributor-Level 10

$π = \frac{n}{v} R T$

$π = \frac{W}{M * v} R T$

$5.03 * 1 0^{- 3} = \frac{2.5}{M * 0.5} * 0.083 * 300$

$M = \frac{2.5 * 0.083 * 300}{5.03 * 0.5} * 1 0^{3} g / m o l$

Molar mass of protein, M = 24751 g/mol

Molar mass of glycine = 75 g/mol

So; number of glycine units = $\frac{24751}{75} = 330$

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

For the given reactions

Sn²+ 2e^- ® Sn

Sn⁴ + 4e^- ® Sn

The electrode potentials are ; $E_{S n^{2 +} / S n}^{o}$ = -0.140V and $E_{S n^{4 +} / S n}^{o} = 0.010 V .$ The magnitude of standard electrode potential for Sn⁴⁺ / Sn²⁺ i.e. $E_{S n^{4 +} / S n^{2 +}}^{o}$ is ___________ * 10^-2 V. (Nearest integer)

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alok kumar singh

Contributor-Level 10

$E_{S n^{2 +} / S n}^{0} = - 0.140 V = E_{2}^{0}$

$E_{S n^{4 +} / S n}^{0} = + 0.010 V = E_{3}^{0}$

Here, $Δ G_{3}^{0} = Δ G_{1}^{0} + Δ G_{2}^{0}$

$- 4 F E_{3}^{0} = - 2 F E_{1}^{0} - 2 F E_{2}^{0}$

$E_{3}^{0} = \frac{E_{1}^{0} + E_{2}^{0}}{2}$

$+ 0.010 = \frac{E_{1}^{0} + (- 0.140)}{2}$

$E_{1}^{0} = E_{S n^{4 +} / S n^{2 +}}^{0} = + 0.16$

= 16 * 10^-2 V

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

For combustion of one mole of magnesium in an open container at 300 K and1 bar pressure, $Δ_{C} H^{Θ} = - 601.70 k J m o l^{- 1},$ the magnitude of change in internal energy for the reaction is__________ kJ. (Nearest integer)

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alok kumar singh

Contributor-Level 10

For combustion of Mg:

Mg (s) + $\frac{1}{2} O_{2} (g) \to M g O (s)$

Here, $Δ n g = - \frac{1}{2}$

Now using

$Δ H = Δ U + Δ n g R T$

-601.7 = $Δ U + (- \frac{1}{2}) * 8.3 * 1 0^{- 3} * 300$

$Δ U = - 600.45 k J$

So; magnitude of $Δ U$ is 600 kJ (the nearest integer).

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

100g of an ideal gas is kept in a cylinder of 146 L volume at 27°C under 1.5 bar pressure. The molar mass of the gas is _________g mol^-1. (Nearest integer)

(Given : R = 0.083 L bar K^-1 mol^-1)

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alok kumar singh

Contributor-Level 10

Using : PV = nRT

1.5 * 416 = n * 0.083 * 300

n = 25mol = $\frac{W}{M}$

25 = $\frac{100}{M}$

So, molar mass, M = 4 g/mol.

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