Chemistry Test 271

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Chemistry Test 271

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Weekly Quiz Competition

Question 1
4 / -1

Electromagnetic radiations having λ = 310 Å are subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photoelectrons with maximum Kinetic energy.

A
0, no emission will occur

B
2.18 × 10⁶ m/s

C
3.09 × 10⁶ m/sec.

D
8.72 × 10⁶ m/s

Solution
Question 2
4 / -1

Select the correct statement(s).

A
The ideal body, which emits and absorbs all frequencies is called a black body.

B
The exact frequency distribution of the emitted radiation from a black body depends upon its temperature.

C
The radiation emitted goes from a lower frequency to higher frequency as the temperature increases.

D
All are correct

Solution

Option A: The ideal body, which emits and absorbs all frequencies is called a black body

Option B: As temperature increases, distribution of frequency (and energy) increases.

Option C: As temperature increases, radiation emitted is shifted to a lower wavelength, i.e. higher frequency.

Hence, options A, B, C, are correct.
Question 3
4 / -1

Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively

A
16 and 4

B
12 and 5

C
12 and 4

D
16 and 5

Solution

Electronic configuration of Cr atom (z = 24) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹

when ℓ = 1, p - subshell,

Numbers of electrons = 12

when ℓ = 2, d - subshell,

Numbers of electrons = 5
Question 4
4 / -1

Which of the following sets of quantum numbers is correct for an electron in 4f orbital ?

A
n = 4, ℓ = 3, m = + 1, s = + ½

B
n = 4, ℓ = 4, m = – 4, s = – ½

C
n = 4, ℓ = 3, m = + 4, s = + ½

D
n = 3, ℓ = 2, m = – 2, s = + ½

Solution

The possible quantum numbers for 4f electron are n = 4, ℓ = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and s =

Of various possiblities only option (a) is possible.
Question 5
4 / -1

Radial probability density in the occupied orbital of a hydrogen atom in the ground state (1s) is given below

A
Radial probability is maximum when (r = a₀)

B
Radial probability is maximum when

C
Radial probability almost falls to zero when

D
Option A and option C are correct

Solution

Radial probability increases as r increase reaches a maximum value when r = a0 (Bohr’s radius) and then falls. When radial probability is very small.

Thus, (a) and (c) are true.
Question 6
4 / -1

25.0 g of FeSO₄.7H₂O was dissolved in water containing dilute H₂SO₄, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO₄ solution for complete oxidation. The percentage of FeSO₄^. 7H₂O in the acid solution is

A
78%

B
98%

C
89%

D
79%

Solution

M.e. of FeSO₄·7H₂O in 25 mL = m.e. of KMnO₄ used = 2 m.e.

M.e. of FeSO₄·7H₂O in 1000 mL = 80 m.e.

Mass of FeSO₄·7H₂O in solution = (80 / 1) × 2787 × (1 / 1000) = 22.24 gm

% of FeSO₄·7H₂O = (22.24 / 25) × 100 = 88.96% ≈ 89%
Question 7
4 / -1

If 6.3 g of NaHCO₃ are added to 15.0 g CH₃COOH solution, the residue is found to weigh 18.0 g. What will be the mass of CO₂ released in the reaction?

A
4.5 g

B
3.3 g

C
2.6 g

D
2.8 g

Solution

The correct answer is Option B.

The chemical reaction will be:

NaHCO₃ + CH₃COOH →CH₃COONa + H₂O + CO₂

molar mass:

NaHCO₃ = 84

CH₃COOH=60

CH₃COONa=82

CO₂ =44

84gNaHCO₃ + 60gCH₃COOH → 82gCH₃COONa + 44gCO₂

Moles of NaHCO₃ = 6.3/84= 0.075

Moles of CH₃COOH = 15/60= 0.25

∴ NaHCO₃ is the limited reagent.

Moles of CO₂ formed = 0.075

Weight of CO₂ = 0.075×44 = 3.3g
Question 8
4 / -1

Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y are

A
40, 30

B
60, 40

C
20, 30

D
30, 20

Solution

Let atomic weight of x = Mx

atomic weight of y = M_y

we know,

mole = weight /atomic weight

a/c to question,

mole of xy₂ = 0.1

so,

0.1 = 10g/( M_x +2M_y)

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05

0.05 = 9/( 3M_x + 2M_y )

3M_x + 2M_y = 9/0.05 = 9× 20 = 180 g ---(2)

solve eqns (1) and (2)

2M_x = 80

M_x = 40g/mol

and M_y = 30g/mole
Question 9
4 / -1

Number of nitrogen atoms present in 1.4 g of N₂

A
6.022 * 10²²

B
6.022 * 10²³

C
6.022 * 10²⁴

D
6.022 * 10²⁵

Solution

Given weight of N₂ gas = 1.4 g

Molar mass of N₂ gas = 28 g

So, mole = given mass/ molar mass

⇒ mole = 1.4/28 = 1/20 mole

Now, number of molecules = mole × avogadro number

⇒ number of molecules = 1/20 × 6.022 × 10²³

⇒ number of molecules = 3.011 × 10²²

Now, we are asked for number of atoms. In N₂, there are 2 atoms, so to obtain number of atoms we will multiply with 2 in number of molecules.

⇒ number of atoms = 2 × 3.011 × 10²²

⇒ number of atoms = 6.022 × 10²²
Question 10
4 / -1

The following equations are balanced atomwise and chargewise.

(i) Cr₂O₇^2- + 8H+ + 2H₂O₂ → 2Cr³+ + 7H₂O + 2O₂

(ii) Cr₂O₇^2- + 8H+ + 5H₂O₂→ 2Cr³⁺ + 9H₂O + 4O₂

(iii) Cr₂O₇^2- + 8H+ + 7H₂O₂→ 2Cr+ + 11H₂O + 5O₂

The precise equationl equations representing the oxidation of H₂O₂ is/are

A
(i) Only

B
(ii) Only

C
(iii) Only

D
All the three

Solution

Cr₂O₇²⁻ + 8H⁺ + 7H₂O₂ → 2Cr³⁺ + 11H₂O + 5O₂

In equation (i), H₂O₂ is oxidized to O₂, and Cr₂O₇²⁻ is reduced to Cr³⁺. This is the correct balanced equation for the oxidation of H₂O₂.

Equations (ii) and (iii) are not stoichiometrically balanced for the redox process.

The answer (A) is correct.