Model paper

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Questions:

Generic 2

11 x 2 = 22

How long would it take a radio wave of frequency, 6 x 10³ s^-1 to travel from Mars to the Earth, a distance of 8 x 10⁷ km?

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 10¹⁵, calculate the energy of the source.

Find energy of each of the photons which have the wavelength of 0.50 \(\overset { \circ }{ A } \)

Find energy of each of the photons which
(i) correspond to light of frequency 3×10¹⁵ Hz.
(ii) have wavelength of 0.50 Å.

Calculate the mass per cent of different elements present in sodium sulphate, (Na₂SO₄).

What will be the molality of the solution contaning 18.25g of HCI gas in 500 g of water?

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

Calculate the percentage composition of the various elements in \({ MgSO }_{ 4 }\)

Calculate the number of gram of oxygen in 0.10 mole of \({ Na }_{ 2 }{ CO }_{ 3 }.{ 10H }_{ 2 }O.\)

10.

An organometallic compound on analysis was found to contain, C = 64.4%, H = 5.5 %, and Fe = 29.9%. Determine its empirical formula (atomic mass of Fe = 56 \(\mu \) )

11.

Using the unit conversion factor, express 1.54mm s^-1 into pm \(\mu \)s^-1.

Multiple Choice Question

19 x 1 = 19

12.

If the value of azimuthal quantum number is 2, what will be the values for magnetic quantum number?

(a)

(b)

(c)

(d)

13.

The maximum number of electrons in a sub-shell is given by the equation _______.

(a)

n²

(b)

2n²

(c)

21-1

(d)

21 + 1

14.

The energy needed to remove a single electron (most loosely bound) from an isolated gaseous atom is called _______.

(a)

ionisation energy

(b)

electronegativity

(c)

kinetic energy

(d)

electron affinity

15.

The outermost electronic configuration of manganese (at. no. = 25) is _______.

(a)

3d⁵4s²

(b)

3d⁶4s¹

(c)

3d⁷4s^o

(d)

3d⁶ 4s²

16.

The orbital with n = 3 and 1 = 2 is _______.

(a)

(b)

(c)

(d)

17.

de Broglie equation is _______.

(a)

\(\lambda =\frac { h }{ mv } \)

(b)

\(\lambda =\frac { hv }{ m } \)

(c)

\(\lambda =\frac { mv }{ h } \)

(d)

\(\lambda =hmv\)

18.

The idea of stationary orbits was first given by _______.

(a)

Rutherford

(b)

J.J. Thomson

(c)

Niels Bohr

(d)

Max Planck

19.

The Balmer series in the spectrum of hydrogen atom falls in _______.

(a)

ultraviolet region

(b)

visible region

(c)

infrared region

(d)

none of these

20.

In a sodium atom (atomic number = 11 and mass number = 23) and the number of neutrons is _______.

(a)

equal to the number of protons

(b)

less than the number of protons

(c)

greater than the number of protons

(d)

none of these

21.

Cathode rays are deflected by _______.

(a)

electric field only

(b)

electric and magnetic field

(c)

magnetic field only

(d)

none of these

22.

The mass of an atom of nitrogen is _______.

(a)

\(\frac { 14 }{ { { 6.023\times 10 }^{ 23 } } } \)

(b)

\(\frac { 28 }{ { { 6.023\times 10 }^{ 23 } } } \)g

(c)

\(\frac { 1 }{ { { 6.023\times 10 }^{ 23 } } } \)g

(d)

14 amu

23.

The number of grams of oxygen in 0.10 mol of Na₂CO₃· 10H₂O is _______.

(a)

20.8 g

(b)

18 g

(c)

108 g

(d)

13 g

24.

The number of significant figures in 0.0101 is _______.

(a)

(b)

(c)

(d)

25.

The empirical formula of sucrose is _______.

(a)

CH₂O

(b)

CHO

(c)

C₁₂H₂₂ O₁₁

(d)

C(H₂O)₂

26.

Which of the following has the highest mass?

(a)

1 g atom of C

(b)

\(\frac { 1 }{ 2 } \)mole of CH₄

(c)

10 mL of water

(d)

3.011 x 10²³atoms of oxygen

27.

12 g of Mg will react completely with an acid to give: _______.

(a)

1 mole of O₂

(b)

\(\frac { 1 }{ 2 } \)mole of H₂

(c)

1 mole of H₂

(d)

2 mole of H₂

28.

How many grams are contained in 1 gram atom of Na?

(a)

13 g

(b)

1 g

(c)

23 g

(d)

\(\frac { 1 }{ 23 } \) g

29.

5.6 litres of oxygen at NTP is equivalent to _______.

(a)

1 mole

(b)

\(\frac { 1 }{ 4 } \)mole

(c)

\(\frac { 1 }{ 8 } \)mole

(d)

\(\frac { 1 }{ 2 } \)mole

30.

One mole of CO₂ contains _______.

(a)

6.02 x 10²³atoms of C

(b)

3 g of CO₂

(c)

6.02 x 10²³atoms of O

(d)

18.1 x 10²³ molecules of CO₂

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Answers Key:

Generic 2

11 x 2 = 22

All radiations in vacuum travel with the same speed, i.e. 3 x 10⁸ ms^-1
Distance to be travelled from Mars to the Earth
= 8 x 10⁷ km = 8 x 10⁷x 10³ m(1 km = 10³m)
Time Taken = \(\frac { 8\times { 10 }^{ 7 }\times { 10 }^{ 3 } }{ 3\times { 10 }^{ 8 } } =2.66\times { 10 }^{ 2 }s=4min43s\)

Ferquency, V = \(\frac { 1 }{ Period } \)= \(\frac { 1 }{ 2\ ns } \) = \(\frac { 1 }{ 2\times 10^{ -9 }s } \)
= 0.5 x 10⁹s^-1
Energy of the source = Energy of 1 photon x number of photons produced
E_source = hv x N
= 6.626 x 10^-34 Js x 0.5 x 10⁹ s^-1 x 2.5 x 10¹⁵
= 8.28 x 10^-10 J

Energy, \(E=\frac { hc }{ \lambda } \)
\(\lambda =0.50\overset { \circ }{ A } \ =0.50\times { 10 }^{ -10 }m\)
\(E=\frac { 6.626\times { 10 }^{ -34 } \ Js\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 0.50\times { 10 }^{ -10 }m } \)
= 39.756 x 10^-16 J
= 3.975 x 10^-15 J

(i) Energy, E = hv
[h = Planck's constant = 6.626 x 10^-34Js v = 3 x 10¹⁵Hz = 3 x 10¹⁵Cps]
Energy, E = 6.626 x 10^-34 Js x 3 x 10¹⁵ s^-1
= 19.878 x 10^-19 J = 1.9878 x 10^-18 J
(ii) Energy, \(E=\frac { hc }{ \lambda } \)
\(\lambda =0.50\overset { \circ }{ A } \ =0.50\times { 10 }^{ -10 }m\)
\(E=\frac { 6.626\times { 10 }^{ -34 } \ Js\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 0.50\times { 10 }^{ -10 }m } \)
= 39.756 x 10^-16 J
= 3.975 x 10^-15 J

Mass percent of an element
= \( \frac{\text{ Mass of that element in the compund} \times 100 }{\text{ Molar mass of } Na_2SO_4}\)
= (2\(\times\)22.99) + 32.06 + (4\(\times\)16.00) = 142.04 g
Mass percent of Sodium
= \(\frac{45.98 \times 100}{142.04}\)
= 32.37
Mass percent of Sulphur
= \(\frac{32.06\times100}{142.04}\)
= 22.57
Mass percent of oxygen
= \(\frac{64\times100}{142.04} \)
= 45.06

Molality is defined as the number of moles of solute present in 1kg of solvent. It is denoted by m.
Thus Molality (m)
\(=\frac { moles\ of\ solute }{ mass\ of\ solvent } \)
Given that, Mass of solvent (H₂O)
= 500g
= 0.5kg
Weight of HCI
= \(1\times 1+1\times 35.5=36.5g\)
Molar of HCI(solute)
\(=\frac { 18.25 }{ 36.5 } =0.5\)
\(m=\frac { 0.5 }{ 0.5 } =1m\)

In the given question 0.9g L^-1 means that 1000 mL solution contains 0.9 g of glugose
Number of moles = 0.9g glucose = \(\frac { 0.9 }{ 180 } \)mol glugose
= \(5\times { 10 }^{ -3 }\)mol glugose.
(where molecular mass of glugose(C₂H₁₂O₆) = \(12\times 6+12\times 1+6\times 16=180u)\)
i.e IL solution contains 0.05 mole glugose or the molarity of glucose is 0.005M.

Molecular mass of \(\mathrm{MgSO}_4\)
=24+32+4 \times 16=120
Mass of \(\mathrm{Mg}=24\)
Mass percentage of \(\mathrm{Mg}=\frac{24}{120} \times 100=20 \%\)
Mass of \(\mathrm{S}=32\)
mass percentage of \(S=\frac{32}{120} \times 100=26.67 \%\)
Mass of \(\mathrm{O}=4 \times 16=64\)
Mass of percentage of \(\mathrm{O}=\frac{64}{120} \times 100=53.33 \%\)

Step 1:
From the formula, we can say that,
1 mol of Na₂CO₃.10H₂O contains 10+3 = 13 moles O atoms
we have 0.1 moles of Na₂CO₃.10H₂O
So, we will have 13 x 0.1 = 1.3mol O atoms
Step 2:
Since, 1 mole O atoms = 6.022 x 10²³atoms
1.3 mol O atoms = 1.3 x 6.022 x 10²³= 0.78 x 10²⁴O atoms.
Hence, the number of oxygen atoms in 0.10 mole of Na₂CO₃.10H₂O are 0.78×10²⁴.

10.

Element	%	Atomic mass	Relative number of moles	Simplest molar ratio	Simplest whole number molar Ratio
C	64.4	12	\(\frac{64.4}{12}=5.36\)	\(\frac{5.36}{0.53}=10.1\)	10
H	5.50	1	\(\frac{5.50}{1}=5.50\)	\(\frac{5.50}{0.53}=10.4\)	10
Fe	29.9	56	\(\frac{29.9}{56}=0.53\)	\(\frac{0.53}{0.53}=1\)	1