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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

If the concentration of glucose (C₆H₁₂O₆) in blood is 0.9 g L^–1, what will be the molarity of glucose in blood?

(i) 5 M

(ii) 50 M

(iii) 0.005 M

(iv) 0.5 M

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

The molar mass of glucose is 180 g mol^- . The molarity (M) is given by the formula:

M = $\frac{c o n c e n t r a t i o n}{m o l a r m a s s}$

On substituting the value in the above equation, the mol can be calculated as

M = $\frac{0.90 g L^{- 1}}{180 g {m o l}^{- 1}}$ = 0.005M

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Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

(i) 4g He

(ii) 46g Na

(iii) 0.40g Ca

(iv) 12g He

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (D)

(A) The number of moles is given by the following formula,

Moles = $\frac{M a s s}{M o l a r m a s s} \dots$ ….(1)

The number of moles of He is calculated by using equation (1) as follows

Moles of O₂= $\frac{4 g}{4 g / m o l}$ =1 mol

The number of atoms can be calculated as, number of moles

$\frac{n u m b e r o f a t o m s}{A v o g a d r o s n u m b e r} \dots$ ….(2)

On substituting the values in the above equation:

1 mol = $\frac{n u m b e r o f a t o m s}{6.022 * 10^{23}}$

Number of atoms = 1 $*$ 6.022 $*$ 10²³

(B) The number of moles of Na is calculated by using equation (1) as follows,

Moles of Na = $\frac{46 g}{23 g / m o l}$ = 2 mol

The number of atoms can be calculated by using equation (2) as follows,

2 mol = $\frac{n u m b e r o f a t o m s}{6.022 * 10^{23}}$

number of atoms= 2 $*$ 6.022 $*$ 10²³

(C) The n

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (D)

(A) The number of moles is given by the following formula,

Moles = $\frac{M a s s}{M o l a r m a s s} \dots$ ….(1)

The number of moles of He is calculated by using equation (1) as follows

Moles of O₂= $\frac{4 g}{4 g / m o l}$ =1 mol

The number of atoms can be calculated as, number of moles

$\frac{n u m b e r o f a t o m s}{A v o g a d r o s n u m b e r} \dots$ ….(2)

On substituting the values in the above equation:

1 mol = $\frac{n u m b e r o f a t o m s}{6.022 * 10^{23}}$

Number of atoms = 1 $*$ 6.022 $*$ 10²³

(B) The number of moles of Na is calculated by using equation (1) as follows,

Moles of Na = $\frac{46 g}{23 g / m o l}$ = 2 mol

The number of atoms can be calculated by using equation (2) as follows,

2 mol = $\frac{n u m b e r o f a t o m s}{6.022 * 10^{23}}$

number of atoms= 2 $*$ 6.022 $*$ 10²³

(C) The number of moles of Ca is calculated by using equation (1) as follows

Moles of Ca= $\frac{0.40 g}{40 g / m o l}$ = 0.01 mol

The number of atoms can be calculated by using equation (2) as follow

0.01 mol = $\frac{n u m b e r o f a t o m s}{6.022 * 10^{23}}$

number of atoms = 0.01 $*$ 6.022 $*$ 10²³

(D) The number of moles of He is calculated by using equation (1) as follows,

Moles of He = $\frac{12 g}{4 g / m o l}$ = 3 mol

The number of atoms can be calculated by using equation (2) as follows,

3 mol= $\frac{n u m b e r o f a t o m s}{6.022 * 10^{23}}$

number of atoms = 3 $*$ 6.022 $*$ 10²³

Thus, 12 g He contains the highest number of atoms.

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Which of the following statements concerning the quantum numbers are correct?

(i) Angular quantum number determines the three dimensional shape of the orbital.

(ii) The principal quantum number determines the orientation and energy of the orbital.

(iii) Magnetic quantum number determines the size of the orbital.

(iv) Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (i)& (iv)

Principal quantum number (n): It represents the size and energy of orbitals.
Azimuthal quantum number (l): It represents the subshell and shape of orbitals.
Magnetic quantum number (ml): It represents the orientation of the orbitals.
Spin quantum number (ms): It represents the direction of spin of electrons in the orbitals.

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?

(i) 1.5 M

(ii) 1.66 M

(iii) 0.017 M

(iv) 1.59 M

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

The relation between molarity and volume is given as,

M₁ V₁= M₂V₂

On substituting the value in the above equation, the political can be calculated as

5M* 500 mL = M₂*1500 mL

M =1.66M

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

In which of the following pairs, the ions are iso-electronic?

(i) Na⁺, Mg²⁺

(ii) Al³⁺, O^–

(iii) Na⁺ , O^2–

(iv) N^3–, Cl^–

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (i) & (iii)

Isoelectronic species are those species that possess the same number of electrons. Here both Na⁺, Mg²⁺ possess 10 electrons and both Na⁺, O^2– possess 10 electrons.

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Which of the following sets of quantum numbers are correct?

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (ii) & (iii)

For the given value of n (principal quantum number) the value of l (Azimuthal quantum number) varies from 0 to n-1. However for the given value of l the m_l (magnetic quantum number) varies from -1 to +1.

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals :

1. (a) n = 3, l = 2, m_l = –2, m_s= $- \frac{1}{2}$

(b) n = 3, l = 2, m_l= –1, m_s= $- \frac{1}{2}$

2. (a) n = 3, l = 1, m_l= 1, m_s = + $\frac{1}{2}$

(b) n = 3, l = 2, m_l= 1, m_s = + $\frac{1}{2}$

3. (a) n = 4, l = 1, m_l = 1, m_s= + $\frac{1}{2}$

(b) n= 3, l = 2, m_l= 1, m_s= + $\frac{1}{2}$

4. (a) n = 3, l = 2, m_l = +2, m_s = - $\frac{1}{2}$

(b) n = 3, l = 2, m_l = +2, m_s = + $\frac{1}{2}$

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (i) & (iv)

In a multielectron atomic system the energy of an electron depends not only on its principal quantum number (shell), but also on its azimuthal quantum number (subshell). Electrons having the same shells and same subshells have the same energy and they are known as degenerate orbitals.

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

(i) 4 mol L^–1

(ii) 20 mol L^–1

(iii) 0.2 mol L^–1

(iv) 2 mol L^–1

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

No. of mole given = $\frac{g i v e n m a s s}{m o l a r m a s s}$

On substituting the value in the above equation, the cal, can be calculated as

no, of mole = $\frac{5.85 g}{58.5 g {m o l}^{- 1}}$ = 0.1 g

The molarity (M) is given by the formula:

M = $\frac{n}{V_{1}}$

On substituting the values in the above equation:

Molarity = $\frac{0.1 m o l 1000 L}{500}$

= 0.2 mol L^-¹

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Identify the pairs which are not of isotopes?

^{(i) 12} ₆ X , ¹³₆Y

^{(ii) 35} ₁₇X, ³⁷ ₁₇ Y

^{(iii) 14}₆X, ¹⁴₇Y

^{(iv) 8}₄ X, ⁸₅Y

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (iii) & (iv)

The isotopes are defined as atoms with identical atomic numbers but different mass numbers are known as isotopes.

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10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

If travelling at same speeds, which of the following matter waves have the shortest wavelength?

(i) Electron

(ii) Alpha particle (He²⁺)

(iii) Neutron

(iv) Proton

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (ii) Alpha particle (He²⁺)

As the wavelength is inversely proportional to the mass of the particles thus the alpha particles would possess the shortest wavelength among the above given option

$λ = \frac{h}{m v}$

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