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The density of a liquid is $1.2 \mathrm{~g} / \mathrm{mL}$. There are 35 drops in $2 \mathrm{~mL}$. The number of molecules in one drop (molar mass of liquid $=70 \mathrm{~g} / \mathrm{mol}$ ) is :

(A) $\left(\frac{1.2}{35}\right) \mathrm{N}_{\mathrm{A}}$

(B) $\frac{1}{12}\left(\frac{1}{35}\right)^{2} \mathrm{~N}_{\mathrm{A}}$

(C) $\frac{1.2}{(35)^{2}} \mathrm{~N}_{\mathrm{A}}$

(D) $1.2 \mathrm{~N}_{\mathrm{A}}$

Solution —

Mass of one drop $=\frac{2}{35} \times 1.2=\frac{2.4}{35} \mathrm{~g}$ Moles of liquid in one drop $=\frac{2.4}{35 \times 70}=\frac{1.2}{35^{2}}$

Moles of liquid in one drop $\therefore$ Molecules $=\frac{1.2}{35^{2}} \times \mathrm{N}_{\text {A }}$

So, The correct option of this question is (C)

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