Question

The number of molecules present in one drop (0.05 ml) of benzene (density = 0.78 g/ml) is nearly equal to $(N_A=6\times {10}^{23})$:

• A. $6\times {10}^{23}$
• B. $3\times {10}^{21}$
• C. $3\times {10}^{20}$
• D. $6\times {10}^{19}$

Answer 1

Answer: (C)

$3\times {10}^{20}$

Mass of $1$ drop of benzene $=0.05\times 0.78=0.039g$

Moles of benzene $=\frac{0.039}{78}=5\times {10}^{-4}$

$1$ mole of benzene contains $6\times {10}^{23}$ molecules.

$\therefore$ Molecules present in $1$ drop $=5\times {10}^{-4}\times 6\times {10}^{23}=3\times {10}^{20}$