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Some Basic Concepts of Chemistry
Mole Concept
9 questions from 8 papers, each with its answer.
- questions
- 9
- papers
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Appears in
Concepts tested
- Comparing number of moles 2
- Number of molecules 2
- Avogadro number and molar mass
- Mass ratio to mole ratio
- Number of atoms
- number of atoms in 1 g of different elements
- Volume of a single molecule
All Questions
2020 1 question
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Which one of the followings has maximum number of atoms?Number of Mg atoms $= \frac{1}{24}N_A$ Number of O atoms $= \frac{1}{32} \times 2 \times N_A = \frac{1}{16}N_A$ Number of Li atoms $= \frac{1}{7}N_A$ Number of Ag atoms $= \frac{1}{108}N_A$ So 1 g of Li has the maximum number of atoms.
2015 AIPMT-I 1 question
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A mixture of gases contains $H_2$ and $O_2$ gases in the ratio of 1:4 (w/w). What is the molar ratio of the two gases in the mixture?Molar masses: $H_2$ = 2 g/mol, $O_2$ = 32 g/mol.
Moles of $H_2$ : $O_2$ $= \frac{1}{2} : \frac{4}{32} = \frac{1}{2} : \frac{1}{8} = 4 : 1$
2015 AIPMT-II 2 questions
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The number of water molecules is maximum in:1 mole of water = $6.02\times10^{23}$ molecules, so 18 moles = $18\times 6.02\times10^{23}$ molecules.
18 g of water is only 1 mole and 1.8 g is 0.1 mole, so 18 moles of water has the maximum number of molecules. -
If Avogadro number $N_A$ is changed from $6.022\times10^{23}$ $mol^{-1}$ to $6.022\times10^{20}$ $mol^{-1}$, this would change:Mass of 1 mol ($6.022\times10^{23}$ atoms) of carbon = 12 g.
With the new $N_A$, mass of 1 mol ($6.022\times10^{20}$ atoms) = $\frac{12\times6.022\times10^{20}}{6.022\times10^{23}} = 12\times10^{-3}$ g.
So the mass of one mole of carbon changes.
2011 AIPMT-MAINS 1 question
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Which has the maximum number of molecules among the following?Number of molecules = moles × $N_A$, so the sample with the most moles has the most molecules.
8 g $H_2$: $\frac{8}{2} = 4$ mol, i.e., $4N_A$ molecules
64 g $SO_2$: $\frac{64}{64} = 1$ mol, i.e., $N_A$ molecules
44 g $CO_2$: $\frac{44}{44} = 1$ mol, i.e., $N_A$ molecules
48 g $O_3$: $\frac{48}{48} = 1$ mol, i.e., $N_A$ molecules
Hence 8 g of $H_2$ has the maximum number of molecules.
2010 AIPMT-PRE 1 question
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The number of atoms in 0.1 mol of a triatomic gas is ($N_A = 6.02\times10^{23}\ mol^{-1}$)Number of molecules in 0.1 mol = $0.1\times N_A = 0.1\times6.02\times10^{23} = 6.02\times10^{22}$
Each molecule of a triatomic gas contains 3 atoms.
Number of atoms = $3\times6.02\times10^{22}$
$= 1.806\times10^{23}$
2008 AIPMT 1 question
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Volume occupied by one molecule of water (density = 1 g cm$^{-3}$) is :Weight of $6.023 \times 10^{23}$ molecules (1 mole) of water = 18 g
Volume occupied by $6.023 \times 10^{23}$ molecules of water (density = 1 g cm$^{-3}$) $= \dfrac{\text{mass}}{\text{density}} = \dfrac{18\ g}{1\ g\ cm^{-3}}$
$= 18$ cm$^3$ or 18 mL
So the volume occupied by one molecule of water $= \dfrac{18}{6.023 \times 10^{23}}$ cm$^3$
$= 2.988 \times 10^{-23}$ cm$^3$
$\approx 3.0 \times 10^{-23}$ cm$^3$
2004 AIPMT 1 question
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The maximum number of molecules is present in:-The number of molecules is proportional to the number of moles, so compare the moles. At STP, 1 mole of gas occupies 22.4 L.
5 L of $N_2$ at STP: $\dfrac{5}{22.4} = 0.223$ mol
0.5 g of $H_2$: $\dfrac{0.5}{2} = 0.25$ mol
10 g of $O_2$: $\dfrac{10}{32} = 0.3125$ mol
15 L of $H_2$ at STP: $\dfrac{15}{22.4} = 0.67$ mol
The largest number of moles is in 15 L of $H_2$, which contains $0.67 \times 6.022 \times 10^{23} \approx 4.0 \times 10^{23}$ molecules.
So the maximum number of molecules is present in 15 L of $H_2$ gas at STP.
2002 AIPMT 1 question
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Which has maximum molecules : -Number of molecules is proportional to the number of moles $= \dfrac{\text{mass}}{\text{molar mass}}$.
7 g $N_2$: $\dfrac{7}{28} = 0.25$ mol
2 g $H_2$: $\dfrac{2}{2} = 1$ mol
16 g $NO_2$: $\dfrac{16}{46} = 0.35$ mol
16 g $O_2$: $\dfrac{16}{32} = 0.5$ mol
So 2 g of $H_2$ (1 mol, $6.022 \times 10^{23}$ molecules) has the maximum number of molecules.