Question

The radius of water molecule having density $1gm{L}^{-1}$ is:

• A. $1.925A^{\circ }$
• B. $73.46A^{\circ }$
• C. $19.25A^{\circ }$
• D. $7.346A^{\circ }$

Answer 1

Answer: (A)

$1.925A^{\circ }$

Volume of $1gm$ water molecules= $1c{m}^3$

Volume of $\frac{1}{18}$ mole of water= ${10}^{-6}{m}^3$

$\frac{1}{18}$ mole of water= $6.023\times {10}^{23}\times \frac{1}{18}$ molecules of water

Volume of $\frac{1}{18}\times 6.023\times {10}^{23}$ molecules= ${10}^{-6}{m}^3$

Hence, Volume of one water molecule= $0.2988\times {10}^{-28}{m}^3$

Let. the radius of water molecule be $r$

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

$\Rightarrow$ $r=0.1925$ $nm=1.925\mathring{A}$