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Chemistry M.C.Q (1 Marks)

MODEL PAPER 6 · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 12 questions.

Questions

M.C.Q (1 Marks)

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

Question 11 Mark
Match the items given in column I with that in column II:
Column IColumn II
(a) Osmotic Pressure.(i) Cottrell's method.
(b) Relative lowering of vapour pressure.(ii) Rast's method.
(c) Elevation in boiling point.(iii) Ostwald and Walker's method.
(d) Depression in freezing point.(iv) Berkeley and Hartley's method.
Answer
(a) (a) (iv), (b) - (iii), (c) - (i), (d) - (ii).
Explanation: (a) (iv), (b) (iii), (c) - (i), (d) - (ii).
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Question 31 Mark
Which of the following alcohols will not undergo oxidation?
Answer
(b) 2-Methylbutan-2-ol
Explanation: 2-Methylbutan-2-ol will not undergo oxidation.
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Question 41 Mark
The slope in the plot of log $\frac{[ R ]_0}{[ R ]}$ vs. time for a first order reaction is
Answer
(c) $\frac{+ k }{2.303}$
Explanation:First order of reaction is $\frac{+ k }{2.303}$
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Question 51 Mark
What is the correct IUPAC name of the given compound?
Image
Answer
(d) 2,2-Dimethylbutanoic acid
Explanation: 2,2-Dimethylbutanoic acid
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Question 61 Mark
When $10 g$ of radioactive isotope is reduced to $1.25 g$ in $12$ years, the half life period of the isotope is
Answer
$(a) 4$ years
Explanation: Radioactivity follow $1$ order kinetics. $k=\frac{2.303}{t} \log \frac{a}{a-x}$
$k=\frac{2.303}{12} \log \frac{10}{1.25}$
$k=0.1919 \log 8$
$k=0.1919 \times \log 2^3$
$k=0.1919 \times 3 \times \log 2=0.575 \times 0.3010=0.173 yr^{-1}$
now,
$t_{1 / 2}=\frac{0.693}{0.173}=4.005 yrs \simeq 4 yrs$
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Question 71 Mark
An organic compound $X$ is oxidized by using acidified $K_2Cr_2O_7.$ The product obtained reacts with Phenyl hydrazine but does not answer the silver mirror test. The possible structure of $X$ is:
Answer
Secondary alcohol on oxidation forms ketone which reacts with hydrazine bus doesn't gives a silver mirror test.
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Question 81 Mark
The major component of starch is:
Answer
(a) amylopectin
Explanation: Starch is a polymer of a-glucose and consists of two components - Amylose and Amylopectin. Amylose is water-soluble component which constitutes about 15-20% of starch. Amylopectin is insoluble in water and constitutes about 80-85% of starch. Thus, Amylopectin is the major component of starch.
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Question 91 Mark
When Benzene diazonium chloride reacts with phenol, it forms a dye. This reaction is called
Answer
(a) Coupling reaction
Explanation: The reaction is known as Coupling reaction.
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Question 111 Mark
The product of oxidation of $I^-$ with $MnO-_4$ in acidic medium is:
Answer
Iodine is liberated from potassium iodide.
$10 I^{-}+2 MnO_4^{-}+16 H^{+} \rightarrow 2 Mn^{2+}+8 H_2 O+5 I_2$
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Question 121 Mark
Major product formed in the following reaction
Image
Answer
(a).
Image
Explanation:Image



Tert-halide with strong base favours elimination reaction not the substitution reaction.
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