ChemistryQ51–55 of 104 questions
Page 2 of 2 · English
51
ChemistryEasyMCQNEET · 2020
Identify the correct statement from the following.
A
$Lithium$ chloride is deliquescent and crystallises as a hydrate,$LiCl \cdot H_{2}O$.
B
The order of hydration enthalpies of alkaline earth cations is $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
C
$Lithium$ and $Magnesium$ show some similarities in their physical properties as they are diagonally placed in the periodic table.
D
$Lithium$ is the softest among all alkali metals.
Solution
(C) $LiCl$ is deliquescent and crystallises from aqueous solution as a hydrate,$LiCl \cdot 2H_{2}O$.
The hydration enthalpies of alkaline earth metal ions decrease with an increase in ionic size down the group.
The correct order of hydration enthalpy is $Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$.
$Lithium$ shows a diagonal relationship with $Magnesium$,hence their physical and chemical properties are similar.
$Lithium$ is the hardest among all alkali metals.
The hydration enthalpies of alkaline earth metal ions decrease with an increase in ionic size down the group.
The correct order of hydration enthalpy is $Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$.
$Lithium$ shows a diagonal relationship with $Magnesium$,hence their physical and chemical properties are similar.
$Lithium$ is the hardest among all alkali metals.
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52
ChemistryEasyMCQNEET · 2020
One mole of carbon atom weighs $12 \; g$. The number of atoms in it is equal to: (Mass of one carbon-$12$ atom is $1.9926 \times 10^{-23} \; g$)
A
$6.022 \times 10^{23}$
B
$1.2 \times 10^{23}$
C
$6.022 \times 10^{22}$
D
$12 \times 10^{22}$
Solution
(A) The number of atoms in one mole of a substance is defined by Avogadro's number $(N_A)$.
Given that the mass of one mole of carbon is $12 \; g$ and the mass of a single carbon-$12$ atom is $1.9926 \times 10^{-23} \; g$.
The number of atoms $= \frac{\text{Total mass}}{\text{Mass of one atom}} = \frac{12 \; g}{1.9926 \times 10^{-23} \; g} \approx 6.022 \times 10^{23}$ atoms.
Thus,the number of atoms in one mole of carbon is $6.022 \times 10^{23}$.
Given that the mass of one mole of carbon is $12 \; g$ and the mass of a single carbon-$12$ atom is $1.9926 \times 10^{-23} \; g$.
The number of atoms $= \frac{\text{Total mass}}{\text{Mass of one atom}} = \frac{12 \; g}{1.9926 \times 10^{-23} \; g} \approx 6.022 \times 10^{23}$ atoms.
Thus,the number of atoms in one mole of carbon is $6.022 \times 10^{23}$.
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53
ChemistryMediumMCQNEET · 2020
Which of the following oxides is amphoteric in nature?
A
$CO_{2}$
B
$SnO_{2}$
C
$SiO_{2}$
D
$GeO_{2}$
Solution
(B) $CO_{2}$ is an acidic oxide.
$SnO_{2}$ is an amphoteric oxide,meaning it reacts with both acids and bases.
$SiO_{2}$ is an acidic oxide.
$GeO_{2}$ is predominantly acidic in nature.
$SnO_{2}$ is an amphoteric oxide,meaning it reacts with both acids and bases.
$SiO_{2}$ is an acidic oxide.
$GeO_{2}$ is predominantly acidic in nature.
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54
ChemistryMediumMCQNEET · 2020
$CH_3CH_2CH=CH_2 \xrightarrow{B_2H_6 / H_2O_2, OH^{-}} Z$. What is $Z?$
A
$CH_3CH_2CH_2CH_3$
B
$CH_3CH_2CH_2CH_2OH$
C
$CH_3CH_2CH_2CHO$
D
$CH_3CH_2CH(OH)CH_3$
Solution
(B) The reaction of $CH_3CH_2CH=CH_2$ with $B_2H_6$ followed by oxidation with $H_2O_2$ in the presence of $OH^-$ is known as hydroboration-oxidation.
This reaction follows anti-Markovnikov's rule,where the $OH$ group is added to the less substituted carbon atom of the double bond.
Thus,the product $Z$ formed is $CH_3CH_2CH_2CH_2OH$ (butan$-1-$ol).
This reaction follows anti-Markovnikov's rule,where the $OH$ group is added to the less substituted carbon atom of the double bond.
Thus,the product $Z$ formed is $CH_3CH_2CH_2CH_2OH$ (butan$-1-$ol).
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55
ChemistryMediumMCQNEET · 2020
How many $(i)$ $sp^{2}$ hybridised carbon atoms and $(ii)$ $\pi$ bonds are present in the following compound?
A
$8, 5$
B
$7, 5$
C
$8, 6$
D
$7, 6$
Solution
(D) The structure of the compound is methyl phenylpropiolate,$C_6H_5-C \equiv C-COOCH_3$.
$(i)$ The $sp^{2}$ hybridised carbon atoms are the $6$ carbon atoms of the benzene ring and the $1$ carbonyl carbon atom $(C=O)$. Total $sp^{2}$ hybridised carbon atoms = $6 + 1 = 7$.
$(ii)$ The $\pi$ bonds are: $3$ from the benzene ring,$2$ from the triple bond $(C \equiv C)$,and $1$ from the carbonyl group $(C=O)$. Total $\pi$ bonds = $3 + 2 + 1 = 6$.
$(i)$ The $sp^{2}$ hybridised carbon atoms are the $6$ carbon atoms of the benzene ring and the $1$ carbonyl carbon atom $(C=O)$. Total $sp^{2}$ hybridised carbon atoms = $6 + 1 = 7$.
$(ii)$ The $\pi$ bonds are: $3$ from the benzene ring,$2$ from the triple bond $(C \equiv C)$,and $1$ from the carbonyl group $(C=O)$. Total $\pi$ bonds = $3 + 2 + 1 = 6$.
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