Question

A researcher conducts electrochemical machining (ECM) on a binary alloy (density 6000 $kg/m^3$) of iron (atomic weight 56. valency 2) and metal p (atomic weight 24, valency 4). Faraday's constant = 96500 coulomb/mole. Volumetric material removal rate of the alloy is 50 $m{m}^3/s$ at a current of 2000 A. The percentage of the metal P in the alloy is closest to

• A. 40
• B. 25
• C. 15
• D. 79

Answer 1

The correct option is B 25

For iron atomic weight,
$A_{Fe}=56,Valency:Z_{Fe}=2$
For metal P, atomic weight
$A_P=24,Valency:Z_P=4$
For Alloy, density,
$\rho =6000kg/m^3$
$=6gm/cc$

$\text{MRR:}Q=50m{m}^3/s$
$=50\times {10}^{3}cc/s$
$MRR=\frac{eI}{F\rho }$
$50\times {10}^3=\frac{e\times 2000}{96500\times 6}$
$e=14.475$

Now, let % of P in alloy = x

$\therefore \frac{1}{e}=\left(\frac{100-x}{100}\right)\frac{Z_{Fe}}{A_{Fe}}+\frac{x}{100}\times \frac{Z_P}{A_P}$

$\Rightarrow \frac{1}{14.475}=\left(\frac{100-x}{100}\right)\frac{2}{56}+\frac{x}{100}\times \frac{4}{24}$

$\Rightarrow x=24.48\%$