Question

Calculate the number of molecules present in a drop of chloroform (CHCl3) weighing 0.0239 g.
(Atomic mass C = 12 u ; H = 1 u ; Cl = 35.5 u)

Answer 1

The number of molecules present in a drop of chloroform weighing 0.0239 grams is 1.2 × 1020 molecules.

Explanation:

• From the above question, it is known that chloroform’s mass is 0.0239 grams.
• So, the molecular mass of chloroform (CHCl3) = atomic mass of C + atomic mass of H + atomic mass of Cl (3) = (1 × 12) + (1 × 1) + (3 × 35.5) = 119.5g
• In 119.5g of CHCl3, the no. of molecules is 6.022 × 1023
• So, the number of molecules in 0.0239g of CHCl3 = 0.0239 × 6.022 × 1023/119.5.  

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