Question

The number of molecules of water in 333 g of $A{l_2}{(S{O_4})_3}.18{H_2}O$ is:
A. $18.0 \times 6.02 \times {10^{23}}$
B. $9.0 \times 6.02 \times {10^{23}}$
C. $18.0$
D. 36.0

Answer 1

Hint: This question is asked from the basic mole concept topic. Firstly find out the number of moles of $A{l_2}{(S{O_4})_3}.18{H_2}O$ and then find out the number of molecules of water. We need to know that in one mole, there are $6.022 \times {10^{23}}$ molecules.

Complete step by step answer:
First of all, we need to find out the number of moles of $A{l_2}{(S{O_4})_3}.18{H_2}O$ present in the given sample. We know that the formula for number of moles is given mass by molar mass. That is:
${\text{Number of moles = }}\dfrac{{({\text{Given mass)}}}}{{({\text{molar mass)}}}}$
Here, the given mass is 333 grams. The molar mass of $A{l_2}{(S{O_4})_3}.18{H_2}O$ will be equal to:
$M = (2 \times 27) + 3 \times (32 + (3 \times 16)) + (18 \times 18) = 666gmo{l^{ - 1}}$
So, number of moles of $A{l_2}{(S{O_4})_3}.18{H_2}O$ will be equal to: $N = \dfrac{{333}}{{666}} = 0.5$ moles.
Then, the number of molecules of water will be equal to: $18 \times 0.5 \times 6.022 \times {10^{23}} = 9 \times 6.02 \times {10^{23}}$
Where $6.022 \times {10^{23}}$ is the Avogadro number and it is the number of molecules present in one mole of a substance.
Answer is $9 \times 6.02 \times {10^{23}}$

Hence the correct option is B.

Additional Information:
Mole is an easy method for expressing the quantity of a substance. Mole is basically a unit which comprises a large number of particles which cannot be counted normally. It is difficult to count the number of atoms or molecules in a certain amount of substance. So, we use this unit in which Avogadro numbers of atoms or molecules are present.

Note:
The number of moles of a molecule may not always be equal to the number of moles of its constituent elements. For example, one mole of water contains Avogadro number of water molecules. However, each water molecule contains two hydrogen atoms and one oxygen atom.