Question

calculate the no. of molecules present in a drop of chloroform weighting 0.0239g

Answer 1

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Answer 2

Answer: $0.0012\times 10^{23}molecules$

Explanation: According to Avogadro's law, 1 mole of every substance contains Avogadro's number of particles $(6.0323\times 10^{23})$

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.0239g}{119g/mol}=0.0002moles$  

1 mole of chloroform molecule $(CHCl_3)$ contains=$6.0323\times 10^{23}molecules$

0.0002 moles chloroform $(CHCl_3)$ molecule contains=$\frac{6.0323\times 10^{23}}{1}\times 0.0002=0.0012\times 10^{23}molecules$