Question

How many water molecules are there in a drop of volume 0.05 mL? (Density of water is 1 g/mL)
a.- $1.67\times {{10}^{21}}$
b.- $1.67\times {{10}^{22}}$
c.- $1.67\times {{10}^{23}}$
d.- $1.67\times {{10}^{24}}$

Answer 1

An attempt to this question can be made by determining the relation between volume, mass and density. Now substitute the values given in the question to obtain mass of the water sample .Use the following formulas to solve the problem:
Mass = $\rho \text{ x V}$
Where,
$\rho$ stands for density
V stands for volume of solution
Moles = $\dfrac{\text{m }}{M}$
where,
m is the given mass of compound
M is the molar mass of compound

Complete Solution :
Given:
Volume of 1 drop of water = 0.05 ml
Density of water = 1 g/ml
Apply general density volume relation to obtain mass of sample.
Mass of 1 drop of water = $\rho \text{ x V}$
= $0.05\text{ x 1}$
= 0.05 g
- We know that the molecular mass of water is 18 g. So, we can count moles of water in a given amount.
Moles of water = $\dfrac{G\text{iven mass }}{Molecular\text{ }mass}$
= $\dfrac{0.05}{18}$
= 0.0028 moles
Count the number of particles from the number of moles obtained above by the Avogadro’s constant.
No. of water molecules = $n\text{ x }{{N}_{A}}$
Where, n is the number of moles
${{N}_{A}}$ is Avogadro constant
= $0.0028\text{ x }6.02\text{ x 1}{{\text{0}}^{23}}$
= $1.67\text{ x 1}{{\text{0}}^{21}}$
From the above statements we can conclude that the number of water molecules present in a drop of volume 0.05 mL water is $1.67\text{ x 1}{{\text{0}}^{21}}$.
So, the correct answer is “Option A”.

Note: It is important to know that the number of particles present in one mole of any substance will be same and equal to the number $6.02\text{ x 1}{{\text{0}}^{23}}$. This number is called Avogadro constant.