Question

Suppose all the electrons of 100 g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positively charged particle. If these two particles are placed 10.0 cm away from each other. Find the force of attraction between them. Compare it with your weight.

Answer 1

Molecular weight of $H_{2}O=2\times 1\times 16=16$

No. of electrons present in on molecular of $H_{2}O=10$

$18$gm of $H_{2}O$ has $6.023\times {10}^{23}$ molecular

$18$gm of $H_{2}O$ has $6.023\times {10}^{23}\times 10$ electrons.

$100$gm of $H_{2}O$ has $\frac{6.023\times {10}^{24}}{18\times 100}$ electrons.

$\therefore$ number of protons= $\frac{6.023\times {10}^{26}}{18}$

Charge of protons $=\frac{1.6\times {10}^{-19}\times 6.023\times {10}^{26}}{18}$

$=\frac{1.6\times 6.023\times {10}^7}{18}c$

Charge of electrons $=\frac{1.6\times 6.023\times {10}^7}{18}c$

$\therefore$ Electrical force =$\frac{9\times {10}^9\left(\frac{1.6\times 6.023\times {10}^7}{18}\right)\times \left(\frac{1.6\times 6.023\times {10}^7}{18}\right)}{{\left(10\times {10}^{-2}\right)}^2}$

$\begin{array}{c}=\frac{8\times 6.023\times 1.6\times 6.023\times {10}^{25}}{18}\\ =2.56\times {10}^{25}N\end{array}$