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

P₄O₆ reacts with water according to equation P₄O₆ + 6H₂O→ 4H₃PO₃. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P₄O₆ in H₂O.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

P₄O₆ + 6H₂O→4H₃PO₃ (i)

Next is neutralization

4 times H₃PO₃ + 2NaOH→Na₂HPO₃ + 2H₂O (ii)

On addition of 2 equations

P₄O₆+8NaOH→4Na₂HPO₃+2H₂O

P₄O₆ ( mol.mass )= (4*31+16*6)=220

Number of moles of P₄O₆ = $? / V$ = $?$

The product formed is neutralized 8 moles of NaOH

P₄O₆ = 8* $?$ = $?$ molNaOH

Molarity of NaOH in 1 litre = 0.1M

Molarity = $\frac{M o l a r M a s s}{G i v e n m a s s}$

Volume = $\frac{1.1}{220}$

= $\frac{1.1}{220}$ * $\frac{8.8}{220}$ = 0.4 L

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10 months ago

Class 12th Probability

A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement, at most one is defective?

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

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Class 12th Probability

A and B are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ , respectively. If the probability of their making a common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer being correct is:

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t t h a t b o t h o f t h e m s o l v e t h e p r o b l e m \\ ∴ P (E_{1}) = \frac{1}{3} * \frac{1}{4} = \frac{1}{12} \\ a n d E_{2} b e t h e e v e n t t h a t b o t h o f t h e m s a m e i n c o r r e c t l y t h e p r o b l e m . \\ P (E_{2}) = (1 - \frac{1}{3}) * (1 - \frac{1}{4}) = \frac{2}{3} * \frac{3}{4} = \frac{1}{2} \\ L e t H b e t h e e v e n t t h a t b o t h o f t h e m g e t t h e s a m e a n s w e r . \\ H e r e, P (H / E_{1}) = 1, P (H / E_{2}) = \frac{1}{20} \\ ∴ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{\frac{1}{12} * 1}{\frac{1}{12} * 1 + \frac{1}{2} * \frac{1}{20}} = \frac{\frac{1}{12}}{\frac{1}{12} + \frac{1}{40}} = \frac{\frac{1}{12}}{\frac{10 + 3}{120}} = \frac{1 / 12}{13 / 120} = \frac{10}{13} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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10 months ago

Class 12th Probability

In a college, 30% of students fail in physics, 25% fail in mathematics, and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is:

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t t h a t t h e s t u d e n t s f a i l s i n P h y s i c s a n d E_{2} b e t h e e v e n t t h a t t h e s t u d e n t s \\ f a i l s i n M a t h e m a t i c s . \\ ∴ P (E_{1}) = \frac{30}{100}, P (E_{2}) = \frac{25}{100} a n d P (E_{1} \cap E_{2}) = \frac{10}{100} \\ P (E_{1} / E_{2}) = \frac{P (E_{1} \cap E_{2})}{P (E_{2})} = \frac{10 / 100}{25 / 100} = \frac{2}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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10 months ago

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

Explain why ozone is thermodynamically less stable than oxygen?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Due its disintegration into oxygen results in the release of heat ΔH is negative) and an increase in entropy (ΔS is positive), ozone is thermodynamically unstable in comparison to oxygen. These two reactions combine to produce a substantial negative Gibbs energy change (ΔG) for its conversion to oxygen.

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10 months ago

Class 12th Probability

Suppose a random variable $X$ follows the binomial distribution with parameters $n$ and $p$ , where $0 < p < 1$ . If $\frac{P (x = r)}{P (x = n - r)}$ is independent of $n$ and $r$ , then $p$ equals:

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

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10 months ago

Class 12th Probability

For the following probability distribution:

$X$ : 1 2 3 4
$P (X)$ : $\frac{1}{10} \frac{1}{5} \frac{3}{10} \frac{2}{5}$
$E (X^{2})$ is equal to:

(a) 3

(b) 5

(c) 7

(d) 10

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \\ E (X^{2}) = \sum_{i = 1}^{n} P_{i} X_{i}^{2} \\ = 1 * \frac{1}{10} + 4 * \frac{1}{5} + 9 * \frac{3}{10} + 16 * \frac{2}{5} \\ = \frac{1}{10} + \frac{4}{5} + \frac{27}{10} + \frac{32}{5} = \frac{28}{10} + \frac{36}{5} = \frac{100}{10} = 10 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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10 months ago

Class 12th Probability

For the following probability distribution:

$X$ : -4 -3 -2 -1 0
$P (X)$ : 0.1 0.2 0.3 0.2 0.2
$E (X)$ is equal to:

(a) 0

(b) -1

(c) -2

(d) -1.8

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = (- 4) (0.1) + (- 3) (0.2) + (- 2) (0.3) + (- 1) (0.2) + 0 (0.2) \\ = - 0.4 - 0.6 - 0.6 - 0.2 = - 1.8 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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10 months ago

Class 12th Probability

The probability distribution of a discrete random variable $X$ is given below:

$X$ : 2 3 4 5
$P (X)$ : $5 / k, 7 / k, 9 / k, 11 / k$
The value of $k$ is:

(a) 8

(b) 16

(c) 32

(d) 48

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \sum_{i = 1}^{n} P (X_{i}) = 1 \\ ∴ \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1 \\ \frac{32}{k} = 1 \Rightarrow k = 32 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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10 months ago

Class 12th Probability

The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons, 4 are swimmers is:

^{(a) 5}C₄ (0.7)⁴ (0.3)

^{(b) 5}C₁ (0.7) (0.3)⁴

^{(c) 5}C₁ (0.7) (0.3)⁴

(d) (0.7)⁴ (0.3)

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

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