The number of water molecules present in a drop of water (volume 0.0018 ml) at room temperature is
A. 1.568 x 103
B. 6.023 x 1019
C. 4.84 x 1017
D. 6.023 x 1023
Answer: Option B
Solution(By Examveda Team)
The number of water molecules present in a drop of water (volume 0.0018 ml) at room temperature is 6.023 x 1019.This Question Belongs to General Knowledge >> Chemistry
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The volume of water is 0.0018 ml.
The density of water is 1g/ml.
The weight of water is volume × density =0.0018 /1 =0.0018 g
No of mole of water = weight of water given/molecular weight = 0.0018/18 = 1× 10 *-4
No of water molecules = no of moles × avogadro's no = 1×10 *-4 × 6.023×10 *-23 = 6.023×10 *19
According to answer, instead of volume, there must be mass of water in grams.
Density of water = 1.
So 1 = mass/volume, 1 = mass/0.0018, mass = 0.0018g no. of mole = weight/molecular
weight = 0.0018/18.
No of water molecules = No of mole*Avogadros no.
= 0.0018/18*6.023*10exp23 = 6.023*10exp19.