Question

(a) Volume of one molecule of $H_{2}O(d=1g{}^{-3})$.
(b) Radius of $H_{2}O$ molecule (assuming if spherical)

Answer 1

As you know, the density of water is $1g/c{m}^2$ and the molecular weight of water is 18g/mol, it contains $6.023\times {10}^{2}3$ molecules of it.

first of all, we have to find the volume of water=$\frac{mass}{density}$=$\frac{18g}{1g/c{m}^3}$=$18c{m}^3$

hence the volume of water=$18c{m}^3$=$18\times {10}^{-6}{m}^3$

1 mole of water=$6.023\times {10}^{23}$ molecules

the volume of $6.023\times {10}^{23}$ molecules of water=$1.8\times {10}^{-5}{m}^3$

the volume of 1 molecule of water=$\frac{1.8}{6.023}\times {10}^{-}{5}^{-}23$

=$0.2988\times {10}^{28}{m}^3$

b.) let radius of water molecules is r.

then, the volume of 1 molecule of water=$\frac{4}{3}\pi r^3$

$\frac{4}{3}\pi r^3$=$0.2988\times {10}^{-28}{m}^3$

$r^3=0.2988\times \frac{3}{4\pi }\times {10}^{-28}$

$r=0.1925\times {10}^{-9}m=0.1925$ nm