calculate the no of molecules present in drop of chlroform
0.0239
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Step-by-step explanation: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.
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