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Chemistry M.C.Q (1 Marks)

MODEL PAPER 5 · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 12 questions.

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M.C.Q (1 Marks)

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12 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
The rate of reaction A + B → Products, is given by the equation r = k[A] [B]. If B is taken in large excess, the order of reaction would be:
Answer
(d) 1
Explanation: r=k[A][B]
Since [B] is in large excess, so its concentration does not change. So the order of the reaction would be 1.
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Question 21 Mark
The unit of rate constant for a first order reaction is
Answer
unit of rate constant for nth order reaction $(k) = (mol\ L^{-1})^{1-n} 8^{-1}$
put, $n=1;$
unit of rate constant for 1st order reaction$(k) = s^{-1}$​​​​​​​
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Question 31 Mark
Match the items of column I with appropriate entries of column II.
Column IColumn II
(a) Non ideal solution with positive deviation(i) $H _2 O (57 \%)+ HI (43 \%)$
(b) Maximum boiling azeotrope(ii) $CHCl _3+\left( CH _3\right)_2 CO$
(c) Non ideal solution with negative deviation(iii) $C _2 H _5 OH (95.4 \%)+ H _2 O (4.6 \%)$
(d) Minimum boiling azeotrope(iv) $CCl _4+ C _6 H _5 CH _3$
Answer
(a) (a)-(iv), (b) - (i), (c) - (ii), (d) - (iii)
Explanation: (a)-(iv), (b) - (i), (c) - (ii), (d) - (iii)
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Question 41 Mark
An organic compound containing oxygen, upon oxidation forms a carboxylic acid as the only organic product with its molecular mass higher by $14$ units. The organic compound is $..........$
Answer
When $\ce{-CH_2OH}$ group is replaced by $\ce{-COOH}$ group then only molecular weight will increase by $14$ units. .
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Question 51 Mark
Williamson synthesis is used to obtain:
Answer
(a) Ether
Explanation: Williamson synthesis is used to obtain ether. For example,
Image
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Question 61 Mark
Which property of transition metals enables them to behave as catalysts?
Answer
(c) Variable oxidation states
Explanation: Variable oxidation states
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Question 71 Mark
A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives:
Answer
(b) Benzyl alcohol and sodium formate
Explanation: They will undergo Cannizzaro reaction as neither benzaldehyde nor formaldehyde has alpha hydrogen. HCHO will be more reactive towards Cannizzaro compared to benzaldehyde because of less steric hindrance. So, OH nucleophile will attack HCHO first, and then the hydride shift from HCHO to benzaldehyde will occur and thus HCHO will oxidize to HCOO ion and benzaldehyde will reduce to benzyl alcohol.
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Question 81 Mark
Fat soluble vitamins are stored in
Answer
(c) Adipose and liver
Explanation: Vitamins which are soluble in fat and oils but insoluble in water are fat soluble vitamins. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues.
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Question 91 Mark
Which of the following alkyl halides will undergo $\ce{SN1}$ reaction most readily?
Answer
$\ce{(CH3)3C-I}$
$\ce{SN1}$ reactions are generally carried out mainly in polar protic solvents $($like $\ce{H2O})$ and they follow first$-$order kinetics.
This means that the rate of reaction depends only on one reactant.
This reaction favours tertiary alkyl halides because of the high stability of the formed carbocation.
The carbocation forms when the molecule is polarized in water to form a carbocation and halide ion attacked by the nucleophile.
The reactivity of the halides are $R-I> R-Br>R-CI>>R-F$.
Hence$, \ce{(CH3)3C-I}$ will undergo the reaction most readily.
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Question 111 Mark
Out of the following, the strongest base in aqueous solution is
Answer
Dimethylamine
$\ce{NH3} <$ primary amine $<$ tertiary amine $<$ secondary amine
This is because :
$i.$ Steric hindrance
The size of an alkyl group is more than that of a hydrogen atom.
So, an alkyl group would hinder the attack of a hydrogen atom, thus decreasing the basicity of the molecule.
So, the more the number of alkyl groups attached, lesser will be its basicity.
$ii.$ Solvation of ions
When amines are dissolved in water, they form protonated amines.
Also, the number of possibilities for hydrogen bonding also increases.
More the number of hydrogen bonding more is the hydration that is released in the process of the formation of hydrogen bonds.
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Question 121 Mark
The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with:
Answer
$(a)\  \ce{SOCl_2}$ in presence of pyridine
Explanation: The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides, or thionyl chloride. Thionyl chloride $\ce{(SOCl_2)}$ is preferred because the other two products $SO_2$ and $\text{HCl}$ are escapable gases. Hence, the reaction gives pure alkyl halides.
$\ce{ROH + SOCI_2 \rightarrow RCI + SO_2 (g) + HCl (g)}$
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