
Which one of the following represents one mole of the substance?
(A) 16 g of ${{N}_{2}}$
(B) 5.6$d{{m}^{3}}$ of ${{H}_{2}}$ at STP
(C) 11.2 $d{{m}^{3}}$of $C{{l}_{2}}$ at STP
(D) 4 g of Helium
Answer
576.3k+ views
Hint: We can define mole as the amount of substance. For converting between grams and moles, we could use the molar mass of the substance and for converting moles and volume of a gas we could use the concept of number of moles present in 22.4 litre of volume of a gas.
Complete step by step answer:
- Let's start with the idea of Avogadro number and concept of moles. Avogadro’s number can be defined as a proportion that relates molar mass on an atomic scale to physical mass on a human scale. It is defined as the number of elementary particles such as molecules, atoms, compounds, etc. per mole of a substance.
- Or in other words, the number of units in one mole of any substance is called Avogadro’s number or Avogadro’s constant. Its value is given as $6.023\times {{10}^{23}}$.
- Another important property of Avogadro’s number is that the mass of one mole of a substance is equal to that substance’s molecular weight. Also, the gram atomic mass of an element is the mass of one mole of that element and the gram molecular mass of a compound refers to the mass of a single mole of the compound.
- The concept of moles can also be defined in terms of volume. That is at standard temperature and pressure (STP) conditions, one mole of any gas( $6.023\times {{10}^{23}}$ particles) occupies a volume of 22.4 litres. Let's calculate the number of moles in given options from above defined concepts of moles.
The number of moles= $\dfrac{Given\text{ }Mass}{Molar\text{ }mass}$
In 16 g of ${{N}_{2}}$,(molar mass=14)
Number of moles = $\dfrac{16g}{14g}$
= 1.142 moles
In 4 g Helium, (molar mass=4)
Number of moles = $\dfrac{4g}{4g}$
= 1 mole
In 5.6$d{{m}^{3}}$ of ${{H}_{2}}$,
Number of moles = $\dfrac{5.6L}{22.4L}$
= 0.25 Moles
In 11.2 $d{{m}^{3}}$of $C{{l}_{2}}$ ,
Number of moles = $\dfrac{11.2L}{22.4L}$
=0.5 Moles
Therefore, the answer is 4g of Helium contains one mole of the substance.
So, the correct answer is “Option D”.
Note: There will be a confusion on calculating the number of moles in volume. If we are given the volume of gas in litres or $d{{m}^{3}}$, then to calculate number of moles present we should divide the given volume by 22.4 and if we are given number of moles then to calculate the volume multiply the number of moles by 22.4.
Complete step by step answer:
- Let's start with the idea of Avogadro number and concept of moles. Avogadro’s number can be defined as a proportion that relates molar mass on an atomic scale to physical mass on a human scale. It is defined as the number of elementary particles such as molecules, atoms, compounds, etc. per mole of a substance.
- Or in other words, the number of units in one mole of any substance is called Avogadro’s number or Avogadro’s constant. Its value is given as $6.023\times {{10}^{23}}$.
- Another important property of Avogadro’s number is that the mass of one mole of a substance is equal to that substance’s molecular weight. Also, the gram atomic mass of an element is the mass of one mole of that element and the gram molecular mass of a compound refers to the mass of a single mole of the compound.
- The concept of moles can also be defined in terms of volume. That is at standard temperature and pressure (STP) conditions, one mole of any gas( $6.023\times {{10}^{23}}$ particles) occupies a volume of 22.4 litres. Let's calculate the number of moles in given options from above defined concepts of moles.
The number of moles= $\dfrac{Given\text{ }Mass}{Molar\text{ }mass}$
In 16 g of ${{N}_{2}}$,(molar mass=14)
Number of moles = $\dfrac{16g}{14g}$
= 1.142 moles
In 4 g Helium, (molar mass=4)
Number of moles = $\dfrac{4g}{4g}$
= 1 mole
In 5.6$d{{m}^{3}}$ of ${{H}_{2}}$,
Number of moles = $\dfrac{5.6L}{22.4L}$
= 0.25 Moles
In 11.2 $d{{m}^{3}}$of $C{{l}_{2}}$ ,
Number of moles = $\dfrac{11.2L}{22.4L}$
=0.5 Moles
Therefore, the answer is 4g of Helium contains one mole of the substance.
So, the correct answer is “Option D”.
Note: There will be a confusion on calculating the number of moles in volume. If we are given the volume of gas in litres or $d{{m}^{3}}$, then to calculate number of moles present we should divide the given volume by 22.4 and if we are given number of moles then to calculate the volume multiply the number of moles by 22.4.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

