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11 months ago

Class 12th Maharashtra Board

I want sample papers for class 12th Maharashtra board 2026.

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Class 12th Chemistry Coordination Compounds

Total number of relatively more stable isomer(s) possible for octahedral complex $[C u {(e n)}_{2} {(S C N)}_{2}]$ will be________________.

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Raj Pandey

Contributor-Level 9

$[C u {(e n)}_{2} {(S C N)}_{2}]$

More stable isomers = 3 (trans isomers)

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11 months ago

Class 12th Chemistry Chemical Kinetics

For the given first order reaction

A ® B

The half life of the reaction is 0.3010 min. The ration of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to________________. (Nearest integer)

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Raj Pandey

Contributor-Level 9

t_1/2 = 0.301 min

t = 2 min

$K = \frac{2.303}{t} l o g (\frac{C o}{C t})$

$\frac{0.693}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$\frac{2.303 * 0.301}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$∴ 2 = l o g (\frac{C o}{C t})$

$\frac{C o}{C t} = 1 0^{2} = 100$

Ans. 100

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Class 12th Ncert Solutions Chemistry Class 12th

Ka for butyric acid (C₃H₇COOH) is 2 * 10^-⁵. The pH of 0.2 M solution of butyric acid is ____________* 10^-¹ (Nearest integer)

[Given log2 = 0.30]

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Raj Pandey

Contributor-Level 9

Ka for C₃H₇COOH = 2 * 10^-⁵

$p K a = - l o g (2 * 1 0^{- 5}) = 5 - l o g 2$

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

$p H = \frac{p K a - l o g C}{2}$

$= \frac{1}{2} (4.7) - \frac{1}{2} l o g (0.2)$

$p H = 27 * 1 0^{- 1}$

$∴$ Ans 27

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Class 12th Chemistry

150g of acetic acid was contaminated with 10.2g ascorbic acid $(C_{6} H_{8} O_{6})$ to lower down its freezing point by (x * 10^-¹)°C. The value of x is _________. (Nearest integer)

(Given K_f = 3.9 K kg mol^-¹; molar mass of ascorbic acid = 176 g mol^-¹]

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Raj Pandey

Contributor-Level 9

$m = \frac{w * 1000}{m o l e c u l a r w t * w_{s o l v e n t}} = \frac{10.2 * 1000}{176 * 150}$

$= \frac{10200}{176 * 150} = 0.386$

$Δ T_{f} = K_{f} * m$

3.9 * 0.386

$∴ x * 1 0^{- 1} = 15.05 * 1 0^{- 1}$

Ans = 15

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Class 12th Chemistry

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0g cm^-³. The number of atoms in 180g of the element is_________* 10²³ (Nearest integer)

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Raj Pandey

Contributor-Level 9

B.C.C structure

a = 300 pm = 300 * 10^-¹² m

d = 6g/cm³

z = 2

$d = \frac{Z * M}{a^{3}}$

$6 = \frac{2 * A}{{(300 * 1 0^{- 10})}^{3}} = \frac{2 * A}{27 * 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 * 6.022 * 10²³

...more

B.C.C structure

a = 300 pm = 300 * 10^-¹² m

d = 6g/cm³

z = 2

$d = \frac{Z * M}{a^{3}}$

$6 = \frac{2 * A}{{(300 * 1 0^{- 10})}^{3}} = \frac{2 * A}{27 * 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

less

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Class 12th Ncert Solutions Chemistry Class 12th

The compound(s) that is (are) removed as slag during the extraction of copper is:

(A) CaO

B) FeO

C) Al₂O₃

(D) ZnO

(E) NiO

Choose the correct answer from the options given below.

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Raj Pandey

Contributor-Level 9

$F e O + \underset{(A c i d i c f l u x)}{S i O_{2}} \to \underset{(S l a g)}{F e S i O_{3}}$

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Class 12th Chemistry

The IUPAC nomenclature of an element with electronic configuration [Rn] 5f¹⁴6d¹7s² is:

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Raj Pandey

Contributor-Level 9

IUPAC nomenclature of element with atomic no. 103 is uniltrium

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Class 12th Ncert Solutions Chemistry Class 12th

Match the List-I with List-II

List-I	List-II
(A) N₂(g) + 3H₂(g) ® 2NH₃(g)	(I) Cu
(B) CO(g) + 3H₂(g) ® CH₄(g) + H₂O(g)	(II) Cu/ZnO – Cr₂O₃
(C) CO(g) + H₂(g) ® HCOH(g)	(III) Fe_xO_y + K₂O + Al₂O₃
(D) CO(g) + 2H₂(g) ® CH₃OH(g)	(IV) Ni

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

Uses of catalyst is very specific for a particular reaction.

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Class 12th Chemistry

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L^-1 is $0.0405 ° C .$ Density of formic acid is 1.05 g mL^-1. The Van't Hoff factor of the formic acid solution is nearly: (Given for water $K_{f} = 1.86 k k g m o l^{- 1}$ )

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Raj Pandey

Contributor-Level 9

Let we take $1 l$ of solution

Mass of solute = Volume * Density

= 0.5 $m l * 1.05 g m / m l$

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i = 1.9$

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