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  4. Radius of water molecule is (assuming it spherical)

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Q. Radius of water molecule is (assuming it spherical)

Some Basic Concepts of Chemistry

Solution:

Density of water molecule $=1 \, g / mL$
$1 \, g =1 \, mL$
Thus, $\frac{1}{18} mol =1 \, mL$
$\frac{1}{18} \times 6.02 \times 10^{23}$ molecules $=1 \, mL$
Thus, volume of $1$ molecule $=\frac{18}{6.02 \times 10^{23}}$
$=3 \times 10^{-23} mL$
Volume of one spherical molecule $=\frac{4}{3} \pi r^3$
$\frac{4}{3} \pi r^3=3 \times 10^{-23} cm ^3$
$\therefore r^3=\frac{3 \times 3 \times 10^{-23}}{4 \pi} cm ^3$
$r^3=7.162 \times 10^{-23} cm ^3$
$r=1.927 \times 10^{-8} cm$
$=1.927 \times 10^{-10} m$
$=1.927 \mathring{A}$
$=\frac{1.927 \times 10 \times 10^{-10}}{10} m$
$=0.1927\, nm$
$=\frac{1.927 \times 100 \times 10^{-10}}{100} m$
$=192.7 \times 10^{-12} m =192.7\, pm$