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11 months ago

Class 12th 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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11 months ago

Class 12th Thermodynamics

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

In case of adiabatic process

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let $a_{1}, a_{2}, \dots a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2}$ $+ \dots + a_{n}$ . If $a_{1} = 1, a_{1} = 300$ and $15 \leq n \leq 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to:

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alok kumar singh

Contributor-Level 10

$x \frac{d y}{d x} - y = x^{2} (x c o s ? x + s i n ? x), x > 0$

$\frac{d y}{d x} - \frac{y}{x} = x (x c o s ? x + s i n ? x) \Rightarrow \frac{d y}{d x} + P y = Q$

So, I.F. $= e^{\int - \frac{1}{x} d x} \frac{1}{| x |} = \frac{1}{x} (x > 0)$

Thus, $\frac{y}{x} = \int \frac{1}{x} (x (x c o s ? x + s i n ? x)) d x$

$\Rightarrow \frac{y}{x} = x s i n ? x + C$

$? y (π) = π \Rightarrow C = 1$

So, $y = x^{2} s i n ? x + x \Rightarrow (y)_{π / 2} = \frac{π^{2}}{4} + \frac{π}{2}$

Also, $\frac{d y}{d x} = x^{2} c o s ? x + 2 x s i n ? x + 1$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - x^{2} s i n ? x + 4 x c o s ? x + 2 s i n ? x$

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11 months ago

Class 12th 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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11 months ago

Class 12th 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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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If the perpendicular bisector of the line segment joining the points $P (1,4)$ and $Q (k, 3)$ has $y$ -intercept equal to -4 , then a value of $k$ is:

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alok kumar singh

Contributor-Level 10

$m a x {n (A), n (B)} \leq n (A \cup B) \leq n (U)$

$\Rightarrow 76 \leq 76 + 63 - x \leq 100$

$\Rightarrow - 63 \leq - x \leq - 39$

$\Rightarrow 63 \geq x \geq 39$

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11 months ago

Class 12th 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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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be

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Payal Gupta

Contributor-Level 10

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to:

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alok kumar singh

Contributor-Level 10

$f (x) = \int_{1}^{3} ? \frac{\sqrt[]{x} d x}{(1 + x)^{2}} = \int_{1}^{\sqrt[]{3}} ? \frac{t . 2 t d t}{{(1 + t^{2})}^{2}}$ (put $\sqrt[]{x} = t$ )

$= {(- \frac{t}{1 + t^{2}})}_{1}^{\sqrt[]{3}} + {({t a n}^{- 1} ? t)}_{1}^{\sqrt[]{3}}$ [Applying by parts]

$\begin{array}{r} = - (\frac{\sqrt[]{3}}{4} - \frac{1}{2}) + \frac{π}{3} - \frac{π}{4} \\ = \frac{1}{2} - \frac{\sqrt[]{3}}{4} + \frac{π}{12} \end{array}$

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11 months ago

Class 12th 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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