Ncert Solutions Chemistry Class 11th

Match List – I with List – II

List – I (salt) List – II (Flame colour wavelength)

(a) LiCl

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Li -> Crimson Red

Na -> Yellow

Rb -> Red violet

Cs -> Blue

VIBGYOR

Wavelength

Red violet > Crimson Red > Yellow > Blue

$↓$ $↓$ $↓$ $↓$

780 nm 670.8 nm 589.2 nm 455.5 nm

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The correct set from the following in which both pairs are in correct order of melting point is:

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MgO has more lattice enthalpy than NaCl

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Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere.

Reason R : Hydrogen is the lightest element

In the light of the above statements, choose the correct answer from the options given below:

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Hydrogen is most abundant element in earth crust and the lightest element. But in troposphere nitrogen gas in most abundant element

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Match List – I and List – II

Choose the correct answer from the options given below:

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Kindly consider the following figure

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The number of compound/s given below which contain/s – COOH group is………….

(Integer answer)

(a) Sulphanilic acid (b) Picric acid

(c) Aspirin &nb

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Please find the below image

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The correct statement about B₂H₆ is:

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alok kumar singh

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Angle of H_t – B – H_t > angle of H_b – B - H_b

Bond angle $\propto % S - C h a r a c t e r \propto \frac{1}{% o f P - c h a r a c t e r}$

$∠ H_{t}$ B H_t is more so % P-Character is less.

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Consider titration of NoOH solution versus 1.25M oxalic acid solution. At the end point following burette readings were obtained.

(i) 4.5mL (ii) 4.5mL

(iii) 4.44mL

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M_eq of NaOH = M_eq H₂C₂O₄

4.44 * N = 1.25 * 2 * 10

$∴ N = \frac{1.25 * 2 * 10}{4.44} = 5.63$

$∴ N = M = 5.63$

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Five moles of an ideal gas at 293K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is……. kJ mol^-1. (Rounded-off to the nearest integer).

[Use R = 8.314 J mol^-1 K^-1]

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If $Δ E = 0$

Q = W

W = -P_ext $(Δ V) = - 4.3 n R T [\frac{1}{P_{2}} - \frac{1}{P_{1}}]$

= $- 4.3 * 5 * 8.314 * 293 [\frac{1}{1.3} - \frac{1}{2.1}]$

= 15347.70 K = 15.3 kJ

Q = 15 kJ

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Copper reduces $N O_{3}^{-}$ into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed $[C u^{2 +}] a n d P_{N O} = P_{N O_{2}}$ ), the HNO₃ concentration at which the thermodynamic tendency for reduction of $N O_{3}^{-} i n t o$ NO and NO₂by copper is same is 10^x M. The value of 2x is……. (rounded-off to the nearest integer).

$[G i v e n, E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{N O_{3}^{-} / N O}^{0} = 0.96 V, E_{N O_{3}^{-} / N O_{2}}^{0} = 0.79 V a n d a t 298 K, \frac{R T}{F} (2.303) = 0.059]$

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$N O_{3}^{-} + 4 H^{+} + 3 e^{-} \to N O + 2 H_{2} O; E_{N O_{3}^{-}}^{0} = 0.96 V$

$L e t [H N O_{3}] = y M$

For same thermodynamic tendency of reduction

$E_{N O_{3}^{-} / N O} = E_{N O_{3}^{-} / N O_{2}}$

$0.96 - \frac{0.059}{3} l o g \frac{p_{N O}}{y^{5}} = 0.79 - 0.059 l o g \frac{p_{N O_{2}}}{y^{3}}$

Assume $(p_{N O} = p_{N O_{2}} = 1 b a r)$

$0.17 = - \frac{0.0591}{1} l o g \frac{p_{N O_{2}}}{y^{3}} + \frac{0.059}{3} l o g \frac{p_{N O}}{y^{5}}$

$= \frac{0.059}{3} [l o g \frac{p_{N O}}{y^{5}} - l o g \frac{{p^{3}}_{N O_{2}}}{y^{9}}] = \frac{0.059}{3} * l o g [\frac{p_{N O}}{y^{5}} * \frac{y^{9}}{{p^{3}}_{N O_{2}}}] = \frac{0.059}{3} l o g y^{4}$

$l o g y^{4} = \frac{0.17 * 3}{0.059} \Rightarrow y = 1 0^{2.16}$

Comparing x = 2.16 and 2x = 4.32

Nearest integer is 4.

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Given below are two statements:

Statement I : The pH of rain water is normally $\sim 5.6$

Statement II : If the pH of rain water drops below 5.6, it is called acid rain.

In the light of the above statements, choose the correct answer from the options given below

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