Question

Calculate the number of molecules present in a drop of chloroform weighing 0.0239 ( C =12 u , H = 1u , chloroform = 35.5 u)

Answer 1

molecuar mass of CHCl3 = 12 + 1 +3 x 35.5 = 119.5g

119.5g contain = 6.022 x 10^23 molecules

thereforce, 0.0239g contain = 6.022 x 10^23 x 0.0239 / 119.5

                                                  = 12.044 x 10^19 molecules 

Answer 2

Answer: The number of molecules in given amount of chloroform is $1.2044\times 10^{20}$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of chloroform = 0.0239 g

Molar mass of chloroform = $[12+1+(3\times 35.5)]=119.5g/mol$

Putting values in above equation, we get:

$\text{Moles of chloroform}=\frac{0.0239g}{119.5g/mol}=2\times 10^{-4}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, $2\times 10^{-4}$ moles of chloroform will contain = $2\times 10^{-4}\times 6.022\times 10^{23}=1.2044\times 10^{20}$ number of molecules

Hence, the number of molecules in given amount of chloroform is $1.2044\times 10^{20}$