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Chemistry Question on Some basic concepts of chemistry

The number of water molecules present in a drop of water weighing 0.018 gm0.018\, gm is

A

6.022×10266.022 \times10^{26}

B

6.022×10236.022 \times10^{23}

C

6.022×10196.022 \times10^{19}

D

6.022×10206.022 \times10^{20}

Answer

6.022×10206.022 \times10^{20}

Explanation

Solution

18 g18\, g of H2O=1 molH _{2} O =1 \,mol of H2O=6.022×1023H _{2} O =6.022 \times 10^{23} molecules

0.018 g0.018\, g of H2O=0.01818 molH _{2} O =\frac{0.018}{18} \,mol of H2OH _{2} O
=0.01818×6.022×1023=\frac{0.018}{18} \times 6.022 \times 10^{23} molecules

=6.022×1020=6.022 \times 10^{20} molecules