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Chemistry M.C.Q (1 Marks)

Model Paper 4 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 12 questions.

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M.C.Q (1 Marks)

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12 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Standard Molar Enthalpy of formation is:
  • A
    the standard enthalpy change for the formation of one mole of a compound from its elements in at a pressure of 10 bar and $30^{\circ} C$.
  • B
    the standard enthalpy change for the formation of one kg of a compound from its elements in their most stable states of aggregation.
  • ✓
    the standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation.
  • D
    the standard enthalpy change for the formation of one mole of a compound from its elements in at a pressure of 2 bar and $25^{\circ}$ C.
Answer
Correct option: C.
the standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation.
(c) the standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation.
Explanation: The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (reference states) is called standard molar enthalpy of formation.
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MCQ 21 Mark
If the concentration of glucose $\left( C _6 H _{12} O _6\right)$ in blood is $0.9 g L { }^{-1}$, what will be the molarity of glucose in blood?
  • A
    50 M
  • ✓
    0.005 M
  • C
    0.5 M
  • D
    5 M
Answer
Correct option: B.
0.005 M
(b) 0.005 M
Explanation: Molarity $=\frac{\text { Concentration in } g / L}{\text { Molar mass }}$
Concentration of glucose in blood $=0.9 g / L$
Molar mass of glucose $\left( C _6 H _{12} O _6\right)=6 \times 12+12 \times 1+6 \times 16=180 g mol ^{-1}$ Molarity $=\frac{0.9}{180}=0.005 M$
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MCQ 31 Mark
Which of the following compounds is/are amphoteric in nature?
  • A
    $As _2 O _3$
  • ✓
    Both $AI _2 O _3$ and $As _2 O _3$
  • C
    $CI _2 O _7$
  • D
    $AI _2 O _3$
Answer
Correct option: B.
Both $AI _2 O _3$ and $As _2 O _3$
(b) Both $AI _2 O _3$ and $As _2 O _3$
Explanation: $A1_2 O _3$ and $As _2 O _3$ are amphoteric in nature. Amphoteric oxides behave as acidic with bases and basic with acids.
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MCQ 41 Mark
On passing vapours of phenol over heated zinc dust it gets reduced to
  • ✓
    benzene
  • B
    toluene
  • C
    $C _6 H _5 OH$
  • D
    aniline
Answer
Correct option: A.
benzene
(a) benzene
Explanation: Phenol is reduced to benzene.
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MCQ 51 Mark
The bond enthalpy depends on?
  • A
    Electronegativity
  • ✓
    All of these
  • C
    Bond length
  • D
    Size of the atom
Answer
Correct option: B.
All of these
(b) All of these
Explanation: The bond enthalpy depends on many factors sizes of atoms involved in the bond, differences in their electronegativity, bond length, electron affinities, etc.
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MCQ 61 Mark
Among the following pairs of orbitals 2 s and $3 s, 4 d$ and $4 f , 3 d$ and 3 p , the orbitals that will experience the larger effective nuclear charge will be:
  • A
    2s, 4d and 3d respectively
  • ✓
    2s, 4d and 3p respectively
  • C
    2s, 4f and 3d respectively
  • D
    3s, 4f and 3d respectively
Answer
Correct option: B.
2s, 4d and 3p respectively
(b) 2s, 4d and 3p respectively
Explanation: An electron is shielded from the attractive interactions of the nucleus by the electrons in the inner shells.The repulsive and attractive interactions of an electron depend on the shell and the orbital in which the electron is present.
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MCQ 71 Mark
Presence of a nitro group in a benzene ring:
  • ✓
    deactivates the ring towards electrophilic substitution.
  • B
    activates the ring towards electrophilic substitution.
  • C
    renders the ring basic.
  • D
    deactivates the ring towards nucleophilic substitution.
Answer
Correct option: A.
deactivates the ring towards electrophilic substitution.
(a) deactivates the ring towards electrophilic substitution.
Explanation: This is because the Nitro is an electron-withdrawing group, it pulls the electron density from the ring towards itself thereby decreasing the electron density in the ring and deactivating the ring towards attack by the electrophile.
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MCQ 81 Mark
de-Broglie equation is
  • A
    $\lambda=\frac{h v}{m}$
  • B
    $\lambda=\frac{m v}{h}$
  • C
    $\lambda= hmv$
  • ✓
    $\lambda=\frac{h}{m v}$
Answer
Correct option: D.
$\lambda=\frac{h}{m v}$
(d) $\lambda=\frac{h}{m v}$
Explanation: Louis de-Broglie proposed that matter, like light, has a dual character.It exhibits wave as well as particle nature. The wavelength of the wave associated with a particle of mass m moving with velocity v is given by $\lambda=\frac{h}{m v}$
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MCQ 91 Mark
In which method oil bath is used?
  • A
    Distillation under reduced pressure
  • B
    Steam distillation
  • C
    Simple distillation
  • ✓
    Fractional distillation
Answer
Correct option: D.
Fractional distillation
(d) Fractional distillation
Explanation: Fractional distillation
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MCQ 101 Mark
Around $10^{15} Hz$ corresponds to the region of the electromagnetic spectrum
  • A
    ultraviolet region
  • B
    infrared region
  • ✓
    visible light
  • D
    microwave region
Answer
Correct option: C.
visible light
(c) visible light
Explanation: Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 700 nm . In terms of frequency, this corresponds to a band in the vicinity of 430-770 THz.
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MCQ 111 Mark
If $V_f$ is the final volume and $V_i$ is the initial volume and $p_{e x}$ external pressure then the work is done can be calculated by?
  • A
    $\int_{V_i}^{V_f} p_{e x} d V$
  • B
    $\int_0^{V_f} p_{e x} d V$
  • ✓
    $-\int_{V_i}^{V_f} p_{e x} d V$
  • D
    $-\int_{V_i}^0 p_{e x} d V$
Answer
Correct option: C.
$-\int_{V_i}^{V_f} p_{e x} d V$
(c) $-\int_{V_i}^{V_f} p_{e x} d V$
Explanation: $-\int_{V_i}^{V_f} p_{e x} d V= w$
It is equal to the integration of change in volume with respect to the initial and final volume.
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MCQ 121 Mark
In the ethylene molecule the two carbon atoms have the oxidation number
  • A
    $-1,-2$
  • B
    $-1,-1$
  • ✓
    $-2,-2$
  • D
    $+2,-2$
Answer
Correct option: C.
$-2,-2$
(c) $-2,-2$
Explanation: In general carbon molecule have 4 valence electrons and in ethylene molecule $\left( H _2 C = CH _2\right)$ each carbon atom is surrounded by 6 valence electron so by calculating its oxidation number we will get $4-6=-2$. so, thats why carbon have $-2,-2$ oxidation number.
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