Question

The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid =70):
(A) $\dfrac{{1.2}}{{35}}{N_A}$
(B) ${\left( {\dfrac{1}{{35}}{N_A}} \right)^2}$
(C) $\dfrac{{1.2}}{{{{(35)}^2}}}{N_A}$
(D) 1.2${N_A}$

Answer 1

Hint: The formula that relates the weight and the volume of the substance is
\[{\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}\]
The number of moles of the substance is the ratio of the weight of the substance to the molecular weight of the substance.

Complete step by step solution:
We are given to find the numbers of molecules in one drop of the liquid. For that, we need to know that weight of one drop of the liquid. We will find that by using the Avogadro number.
- It is given that there are 35 drops in 2 mL of the liquid. So, we can say that the volume of 1 drop of the liquid will be $\dfrac{2}{{35}}mL$.
- Now, we will find the weight of a drop of a liquid using the density of the liquid. We know that
\[{\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}\]
We are given that density is 1.2 g/mL. So, putting that in the above equation, we get
\[1.2 = \dfrac{{{\text{Weight}} \times {\text{35}}}}{2}\]
So,
\[{\text{Weight = }}\dfrac{{1.2 \times 2}}{{35}}g\]
Thus, we get the weight of a drop of the liquid. Now, we will use the fact that 1 mole of a substance will contain ${N_A}$ number of molecules.
- We are given that the molecular weight of the liquid is $70$ $gmmo{l^{ - 1}}$.
So, the number of moles of liquid present in a drop = $\dfrac{{{\text{Weight}}}}{{{\text{Molecular weight}}}} = \dfrac{{1.2 \times 2}}{{35 \times 70}} = \dfrac{{1.2}}{{{{(35)}^2}}}$
Now, 1 mole of a liquid contains ${N_A}$ number of molecules, so, $\dfrac{{1.2}}{{{{(35)}^2}}}$ moles will contain $\dfrac{{1.2}}{{{{(35)}^2}}}{N_A}$ number of molecules.

Thus, we can say that the correct answer is (C).

Note: Here, the answer in the options is given in a way that it includes ${N_A}$ which is Avogadro's number. So, we do not need to put the value of this number in the calculation and find the absolute value, instead, we need the answer that contains ${N_A}$. The value of ${N_A}$ is $6.022 \times {10^{23}}$.