Question

The volume of a drop of water is 0.0018 ml then number of water molecules present in two drop of water at room temperature is :

• A. $12.046\times {10}^{19}$
• B. $1.084\times {10}^{18}$
• C. $4.84\times {10}^{17}$
• D. $6.023\times {10}^{23}$

Answer 1

The correct option is A $12.046\times {10}^{19}$

$Density=\frac{mass}{volume}$
⇒$mass=density\times volume$
Density of water =1g/ml
Volume of water =0.0018mL(Given)
∴ Mass of 0.0018mL water =$1\times 0.0018=0.0018gm$
Molar mass of water =18gm/mol
Number of moles in 0.0018gm of water =$\frac{Givenmass}{Molarmass}=\frac{0.0018}{18}=0.0001$ mole
As we know that,
Number of molecules of water in 1 mole =$6.022\times {10}^{23}$ molecules
∴ Number of molecules of water in 0.0001 mole =$6.022\times {10}^3\times 0.0001=6.023\times {10}^{19}$
Hence, the number of water molecules present in a 2 drops of water (volume =0.0018ml) at room temperature is $12.046\times {10}^{19}$. {Option A }