Chemistry

This is a Short Answer Type Questions as classified in NCERT Exemplar

Mass of NaoH = 40 g

Mass of solvent =1000 g

Mass of solution = 40 x 3+1000

Density = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}$

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$ = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{D}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}}$

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$ = $\frac{1120\mathrm{g}}{1.10\mathrm{g}/\mathrm{m}\mathrm{L}}$= 1009.0 mL

Molarity = $\frac{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}}$  = $\frac{3}{1.009\mathrm{}}=$  2.97M

1009.00 mL= 1.009 L

Hence, the molarity of the solution is 2.97M

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: (ii) Heisenberg's uncertainty principle.

According to Heisenberg's uncertainty principle it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Thus it implies that determining the trajectory of an electron is impossible as it requires exact position and velocity which is not possible as per the uncertainty principle.

This is a Short Answer Type Questions as classified in NCERT Exemplar

65.3 g of Zinc gives 22.7 litres of Hydrogen gas

32.65 g Zinc gives = 32.65g x 22.7 litres/65.3 = 11.35 L

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (iii) 2

The number of angular nodes is given by n-l

where n is principal quantum number, l is azimuthal quantum number

For 4d orbital, n=4 and l=2

Thus, the number of angular nodes

 = n-l

 = 4-2

 = 2

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  option (iv)

The number of radial nodes is given by n-l-1

where n is principal quantum number, l is azimuthal quantum number

For 3p orbital, n=3 and l=1

Thus, the number of radial nodes

= n-l-1 

= 3-1-1

 = 1

This is a Short Answer Type Questions as classified in NCERT Exemplar

(Natural abundance of 1H x molar mass ) + (Natural abundance of ²H x molar mass of 2H)

Natural abundance of  ¹H = 99.985

Natural abundance of ²H = 0.015

Average atomic mass = $\frac{99.985+0.030}{100\mathrm{}}$

$\frac{100.015}{100\mathrm{}}$ = 1.00015u

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  Option (iv)

sum of the number of protons and neutrons is same but the number of protons is different

Isobars are the atoms which have same mass number (sum of number of neutrons and protons) but different atomic number (i.e. different proton number)

For example, ₁₄⁶C and ₁₄⁷N.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  Option (i)

Overall neutrality of atom.

According to the J.J Thomson model the positive charge is uniformly distributed and the electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement just as watermelon of positive charge with plums or seeds (electrons) embedded into it. Thus this model is able to explain the overall neutrality of the atom

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  Option (ii)

The mass of an electron is equal to the mass of a neutron.

The neutron is heavier than the electron as

Mass of the neutron = 1.67 x 10^-27 kg

Mass of the electron = 9.11 x 10^-31kg