Question

Radius of water molecule is (assuming it spherical)

• A. $19.27nm$
• B. $19.27\mathring{A}$
• C. $192.7pm$
• D. $19.27\mu m$

Answer 1

Answer: (C)

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