
Which of the following weighs the least?
a. $2$g atom of N (atomic weight of N = 14)
b. $3 \times {10^{23}}$ atoms of C (atomic weight of C = 12)
c. 1 mole of S (atomic weight of S = 32)
d. 7 g silver (atomic weight of Ag = 108)
Answer
558k+ views
Hint:We are asked to find out the least weight, take the S.I unit of weight (kilogram) to determine the answer. Hence, convert all given values to kilogram (Kg).
Formula used: Numbers of moles = Given mass/ Molar mass
Complete answer:
a. Let’s first convert the given 2 g atom of Nitrogen(N),
2g atoms of nitrogen = 2 moles of N,
We know 1 mole of N is of 14 g, therefore 2 moles of N will have,
$ \Rightarrow 2 \times 14 = 28{\text{g}}$
Now, because 1kg = 1000g, we can conclude that,
$1$g = $\dfrac{1}{{1000}}$ kg
Therefore, 28 g of nitrogen must be equal to, $\dfrac{1}{{1000}} \times 28$ kg of nitrogen.
$ \Rightarrow\dfrac{1}{{1000}} \times 28$ kg = \[0.028\] kg of nitrogen.
b. Now, it is time to convert $3 \times {10^{23}}$ atoms into kg
But first, we need to change the given atoms into weight by using the aforementioned formula of the number of moles. We know,
1 mole of any atom contains $6.022 \times {10^{23}}$ numbers of atoms,
1 mole = $6.022 \times {10^{23}}$ atoms, therefore
$ \Rightarrow $ 1 atom = $\dfrac{1}{{6.022 \times {{10}^{23}}}}$ moles
We are given $3 \times {10^{23}}$ atoms of carbon that means
$3 \times {10^{23}}$ atoms of carbon = $\dfrac{1}{{6.022 \times {{10}^{23}}}}$ $ \times $ $3 \times {10^{23}}$ moles of carbon.
= 0.498 moles of carbon
Now, using the formula for the number of moles, we can find out the weight of $3 \times {10^{23}}$ atoms of carbon, i.e.,
No. of mole = Given mass OR required mass/ Molar mass
$ \Rightarrow $ Required mass = no. of moles $ \times $ molar mass
Also, the molar mass of a carbon atom is $12$g,
$ \Rightarrow $0.498$ \times \dfrac{{12}}{{1000}}$ = 0.0059 kg of carbon. (since we are calculating in terms of a kilogram)
c. We have 1 mole of sulfur, since it is given in, no. of moles we can directly use the formula,
We know,
No. of mole = Given mass OR required mass/ Molar mass
$ \Rightarrow $ Required mass = no. of moles $ \times $ molar mass
Also, the molar mass of sulfur(S) is 32g
$ \Rightarrow $1$ \times \dfrac{{32}}{{1000}}$ = 0.032kg of sulfur.
d. In this case, we just have to convert the given mass from gram to kilogram
Therefore, 7 grams of silver will be equal to
$ \Rightarrow $$\dfrac{1}{{1000}}$$ \times $ 7 = 0.007 kg of silver.
Now you can easily identify the option with the least weight. i.e., option (b) $3 \times {10^{23}}$ atoms of C.
Note:
Even though we have used the S.I unit of mass Kg, but since all the options in this question had a similar unit ie., g, we would still have got the same results. However, if anyone in the unit did differ from the others then we have to change the unit. It is advised to measure all the quantities in S.I units.
Formula used: Numbers of moles = Given mass/ Molar mass
Complete answer:
a. Let’s first convert the given 2 g atom of Nitrogen(N),
2g atoms of nitrogen = 2 moles of N,
We know 1 mole of N is of 14 g, therefore 2 moles of N will have,
$ \Rightarrow 2 \times 14 = 28{\text{g}}$
Now, because 1kg = 1000g, we can conclude that,
$1$g = $\dfrac{1}{{1000}}$ kg
Therefore, 28 g of nitrogen must be equal to, $\dfrac{1}{{1000}} \times 28$ kg of nitrogen.
$ \Rightarrow\dfrac{1}{{1000}} \times 28$ kg = \[0.028\] kg of nitrogen.
b. Now, it is time to convert $3 \times {10^{23}}$ atoms into kg
But first, we need to change the given atoms into weight by using the aforementioned formula of the number of moles. We know,
1 mole of any atom contains $6.022 \times {10^{23}}$ numbers of atoms,
1 mole = $6.022 \times {10^{23}}$ atoms, therefore
$ \Rightarrow $ 1 atom = $\dfrac{1}{{6.022 \times {{10}^{23}}}}$ moles
We are given $3 \times {10^{23}}$ atoms of carbon that means
$3 \times {10^{23}}$ atoms of carbon = $\dfrac{1}{{6.022 \times {{10}^{23}}}}$ $ \times $ $3 \times {10^{23}}$ moles of carbon.
= 0.498 moles of carbon
Now, using the formula for the number of moles, we can find out the weight of $3 \times {10^{23}}$ atoms of carbon, i.e.,
No. of mole = Given mass OR required mass/ Molar mass
$ \Rightarrow $ Required mass = no. of moles $ \times $ molar mass
Also, the molar mass of a carbon atom is $12$g,
$ \Rightarrow $0.498$ \times \dfrac{{12}}{{1000}}$ = 0.0059 kg of carbon. (since we are calculating in terms of a kilogram)
c. We have 1 mole of sulfur, since it is given in, no. of moles we can directly use the formula,
We know,
No. of mole = Given mass OR required mass/ Molar mass
$ \Rightarrow $ Required mass = no. of moles $ \times $ molar mass
Also, the molar mass of sulfur(S) is 32g
$ \Rightarrow $1$ \times \dfrac{{32}}{{1000}}$ = 0.032kg of sulfur.
d. In this case, we just have to convert the given mass from gram to kilogram
Therefore, 7 grams of silver will be equal to
$ \Rightarrow $$\dfrac{1}{{1000}}$$ \times $ 7 = 0.007 kg of silver.
Now you can easily identify the option with the least weight. i.e., option (b) $3 \times {10^{23}}$ atoms of C.
Note:
Even though we have used the S.I unit of mass Kg, but since all the options in this question had a similar unit ie., g, we would still have got the same results. However, if anyone in the unit did differ from the others then we have to change the unit. It is advised to measure all the quantities in S.I units.
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

