Chemistry Test 131

Result

Chemistry Test 131

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

The intermediate formed in the following reaction is-

A
Carbene

B
Nitrene

C
benzyne

D
Carbonium ion

Solution

The reaction is an example of cine subsitution or elimination addition
Question 2
4 / -1

The compound present in brorax bead is

A
B₂O₃

B
NaBO₂

C
NaBO₃

D
NaBO₂ + B2O₃

Solution

This test is meant for detecting the coloured cations. When borax is heated on a platinum loop it swells up into a white porous mass first which on subsequent heating melts into a shining transparent glass bead (B₂O₃).

When this glass bead is heated with coloured compounds, a characteristic colour of metaborate is formed.

However, in a reducing flame, a different colour is obtained.
Question 3
4 / -1

The correct decreasing order of acidic nature is

A
HCIO > HBrO > HIO

B
HIO > HBrO > HCIO

C
HBrO > HIO > HCIO

D
HCIO > HIO > HBrO

Solution

Acidic character depends on stability of conjugate base. More stable conjugate base, more will be the acidic character

Electronegativity of Cl > Br > I So stability order of conjugate bases CIO- > BrO- > IO- Hence acidic character HCIO > HBrO > HIO
Question 4
4 / -1

In the formal charge on each oxygen atom and P - 0 bond order respectively are

A
- 0.75, 0.0

B
-0.75,1.0

C
-0.75,1.25

D
-3,1.25

Solution

3 negative charges of the ion is beings h are d by 4 oxygen atoms,
∴ formal charge on an O atom = -3/4
Question 5
4 / -1

Which one of the following compounds is not aromatic?

A

B

C

D

Solution

Aromaticity of a compound can be decided by Huckel's rule.
Huckel's rules are as follows:
1. Compound must be cyclic
2. The cyclic compound must have resonance and complete delocalization of electrons i.e., there must be conjugation.
3. The total number of pi electrons must fit the formula = 4n + 2, where n is a non - negative integer.
The important summary of this concept is as follows: (Remember this summary)
Question 6
4 / -1

The actinoids exhibit more number of oxidation states in general than the lanthanoids because

A
The 5f orbitals extend further from the nucleus than the 4f orbitals

B
The 5f orbitals are more buried than the 4f orbitals

C
There is a similarity between 4f and 5f orbitals in their angular part of the wave function.

D
The actinoids are more reactive than lanthanoids

Solution

shielding effect of 5f < shielding effect of 4f

so Z_eff (z effect) in actinoidis < Zeff (z effect) in lanthanoid

so extraction of electrons in actinoid is easy than lanthanoid which has 4f and better shielding.
Question 7
4 / -1

The compound when kept in pentane converts to a constitutional isomer. Identify that
isomer.

A

B

C

D

Solution
Question 8
4 / -1

The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)

A
4.2 x 10^-4 mole

B
4.2 x 10^-6 mole

C
3.82 X 10^-4 mole

D
3.82 X 10-⁶ mole

Solution
Question 9
4 / -1

0.400 g of impure CaSO₄ (Molar mass = 136) solution when treated with excess of barium chloride solution, gave 0.617 g of anhydrous BaSO⁴ (Molar mass = 233). The percentage of CaSO⁴ present in the sample is

A
80.0%

B
90.0%

C
95.0%

D
75.0%

Solution
Question 10
4 / -1

At temperature 300 K for reaction C₂H₄(g) + 3O₂(g)→ 2CO₂(g) + 2H₂O(l); ΔH = − 336.2 kcal. The approximate value of ΔU at 300 K for the same reaction (R = 2 cal degree^{− 1} mol^{− 1}) will be-

A
- 320.0 kcal

B
- 335.0 kcal

C
- 337.2 kcal

D
- 353.0 kcal

Solution

C₂H₄(g) + 3 O₂(g)→2 CO₂(g) + 2 H₂O(l)

At 300 K ΔH = − 336.2 kcal

Δn_g = 2 − (1 + 3) = − 2

T = 300 K, ΔH = − 336.2 kcal,

R = 2 × 10⁻³ kcal degree⁻¹ mol⁻¹

ΔH = ΔU+ Δn_gRT

− 336.2 = ΔU + (− 2) × 300 × 2 × 10⁻³

ΔU = − 336.2 + 1.2 = − 335.0 kcal