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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

When 600 mL of 0.2 M HNO₃ is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is_________ * 10^-² °C.

(Enthalpy of neutralization = 57 kJ mol^-¹ and Specific heat of water = 4.2 JK^-¹ g^-¹)

(Neglect heat capacity of flask)

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Vishal Baghel

Contributor-Level 10

HNO₃ + NaOH ® NaNO₃ + H₂O

t = 0, (600 * 0.2)_meq (400 * 0.1)_meq 40 meq 40 meq

120 meq 40 meq

t = $\infty$ 80 meq -

Total heat

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HNO₃ + NaOH ® NaNO₃ + H₂O

t = 0, (600 * 0.2)_meq (400 * 0.1)_meq 40 meq 40 meq

120 meq 40 meq

t = $\infty$ 80 meq -

Total heat produced during neutralization ;

$= \frac{40}{1000} * 57 * 1000 J = 2280 J = m . s θ$

2280 = 1000 * 4.2 * q

$∴ θ = 2280 / 4200 = 54.28 * 1 0^{- 2} C$

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

The minimum uncertainty in the speed of an electron in an one dimensional region of length 2a₀ (Where a₀ = Bohr radius 52.9 pm) is__________ km s^-¹.

(Given: Mass of electron = 9.1 * 10^-³¹ kg, Planck's constant h = 6.63 * 10^-³⁴ Js)

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Vishal Baghel

Contributor-Level 10

According to Heisenberg's uncertainly principle

$Δ x * Δ p = \frac{h}{4 π}$

$∴ Δ v = \frac{6.63 * 1 0^{- 34}}{4 * 3.14 * 2 * 52.9 * 1 0^{- 12} * 9.1 * 1 0^{- 31}}$

= 548273 ms^-1

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

Ionic radii of cation A+ and anion B are 102 and 181 pm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for B. A+ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is________ pm.

(Nearest integer)

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Vishal Baghel

Contributor-Level 10

Edge length of the unit cell AB = a = 2 $(r_{+} + r_{-})$

$∴ a = 2 (102 + 181) = 2 (283) = 566$

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Three

Resistance of a conductivity cell (cell constant 129 m1) filled with 74.5 ppm solution of KCl is $100 Ω$ (labeled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is $50 Ω$ (labeled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $\frac{\land_{1}}{\land_{2}} = x * 1 0^{- 3} .$ The value of x is________. (Nearest integer)

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Vishal Baghel

Contributor-Level 10

Given cell constant $\frac{l}{A} = 129 m^{- 1}$

Ratio of ppm of two solution of KCl = Ratio of moles of per unit volume.

$\frac{{(p p m)}_{K C l_{1}}}{{(p p m)}_{K C l_{2}}} = \frac{M_{1}}{M_{2}} = \frac{74.5}{149.0} = \frac{1}{2}$

$∴ \frac{\land_{1}}{\land_{2}} = 1000 * 1 0^{- 3}$

$∴$ x = 1000

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Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Sixteen

Which of the following compounds is an example of hypnotic drug?

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Vishal Baghel

Contributor-Level 10

Drug used in the treatment of sleeping problems

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

Which among the following is the strongest Bronsted base?

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Vishal Baghel

Contributor-Level 10

Due the absence of inversion and non resonating nature

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Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Sixteen

Which among the following pairs has only herbicides?

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Vishal Baghel

Contributor-Level 10

The pair of herbicides is sodium chlorate and sodium arsinite

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Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

In neutral of faintly alkaline medium KMnO₄ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be:

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Vishal Baghel

Contributor-Level 10

Following is the reaction of KMnO4 in basic medium with thiosulphate anion.

$\overset{+ 7}{8 M n O_{4}^{-}} + 3 S_{2} O_{3}^{- -} + H_{2} O \to \overset{+ 4}{8 M n O_{2 |}} + 6 S O_{4}^{- -} + 2 O H^{-}$

Change = 7 – 4 = 3

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

In following pairs, the one in which both transition metal ions are colourless is:

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Vishal Baghel

Contributor-Level 10

$S c^{3 +} - 3 d^{0}$ electron in last subshell

$Z n^{2 +} - 3 d^{10}$ electron in last subshell

$T i^{4 +} - 3 d^{0}$ electron in last subshell

$C u^{2 +} - 3 d^{9}$ electron in last subshell

$V^{2 +} - 3 d^{3}$ electron in last subshell

$T i^{3 +} - 3 d^{1}$ electron in last subshell

$M n^{2 +} - 3 d^{5}$ electron in last subshell

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

Number of lone pairs of electrons in the central atom of SCl₂, O₃, ClF₃ and SF₆, respectively are:

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Vishal Baghel

Contributor-Level 10

SCl₂ – 2 lone pair of central atom

O₃ – 1 lone pair of central atom

ClF₃ – 2 lone pair of central atom

SF₆ – 0 lone pair of central atom

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