Question

A potential of $12.0$ volts was applied to two electrodes placed $20cm$ apart. A dilute solution of ammonium chloride was placed between the electrodes when $N{H}_4^{+}$ ions was found to cover a distance of $1.60cm$ in one hour. What is the mobility of $N{H}_4^{+}$ ion?

• A. $7.41\times {10}^{-8}{m}^2{V}^{-1}{s}^{-1}$
• B. $11.2\times {10}^{-8}{m}^2{V}^{-1}{s}^{-1}$
• C. $2.34\times {10}^{-8}{m}^2{V}^{-1}{s}^{-1}$
• D. $21.45\times {10}^{-8}{m}^2{V}^{-1}{s}^{-1}$

Answer 1

The correct option is A $7.41\times {10}^{-8}{m}^2{V}^{-1}{s}^{-1}$

$\text{Distance travelled by}N{H}_4^{+}\text{ion in 1 hour}=1.60\times {10}^{-2}m$

$\therefore$
$\text{Distance travelled by}N{H}_4^{+}\text{ion per second}=\frac{1.60\times {10}^{-2}}{3600}m$

$\text{Potential gradient}=\frac{\text{Voltage}}{\text{Distance between electrodes}}$

$\text{Potential gradient}=\frac{12}{20}Vc{m}^{-1}$

$\text{Potential gradient}=\frac{12\times 100}{20}V{m}^{-1}$

$\therefore$
$\text{Mobility of}N{H}_4^{+}\text{ions}=\frac{\text{Speed}}{\text{Potential gradient}}$

$\text{Mobility of}N{H}_4^{+}\text{ions}=\frac{(1.60\times {10}^{-2}\times 20)m{s}^{-1}}{(12\times 100\times 3600)V{m}^{-1}}$

$\text{Mobility of}N{H}_4^{+}\text{ions}=7.41\times {10}^{-8}{m}^2{V}^{-1}{s}^{-1}$