Question

There is going to be a big fight between Popeye and Bluto in few minutes. Popeye gets in energetic mode if his blood contains $0.1$ mg/l concentration of Iron. Due to certain health issues, the absorption of iron in his bloodstream follows rate law: $-\frac{d\left[F{e}_{spinach}\right]}{dt}=k\left[F{e}_{spinach}\right]$ , which earlier used to be a fast absorption. What is the minimum time he should have spinach before the fight so that he defeats Pluto? Assume initial iron concentration in blood to be negligible.
Data given:
Volume of blood in popeye : $5$ litre; Volume of spinach = $1$ litre; Molecular Weight of Fe : $56$ g/mole; Density of spinach = $1$ kg/l; k = $2.31\times {10}^{-2}$ $mi{n}^{-1}$ ; $\frac{\text{Weight of Fe}}{\text{Weight of spinach}}$ = ${10}^{-3}\frac{\text{g of Fe}}{\text{Kg of spinach}}$

• A. $15$ min
• B. $30$ min
• C. $45$ min
• D. $60$ min

Answer 1

Answer: (B)

$30$ min

$F{e}_{spinach}⟶F{e}_{blood}$

Mole initially $N_{A_o}$ $0$

Moles at any time (t) $N_{A_o}=N_{A_o}–a$ $a$

Our time t is the minimum time to have spinach before the fight.

We know the final concentration in blood we want $=0.1$ mg/l

So number of moles ‘a’ equals = $0.1\times {10}^{-3}\frac{g}{l}\times 5l\times \frac{1}{56\frac{g}{mole}}=\frac{(0.5\times {10}^{-3})}{56}$ moles

We know that ${10}^{-3}$ $Fe$ is present in $1$ kg of spinach.

So by looking at the density we can say ${10}^{-3}gFe$ is present in $1$ litre of spinach. Since there is 1 litre of spinach, so we have ${10}^{-3}gFe$ present.

$\therefore N_{A_o}=\frac{{10}^{-3}}{56}$ moles

So, $N_A=N_{A_o}–a=\frac{(0.5\times {10}^{-3})}{56}$ moles

Volume terms can cancel out from rate equation to give :

$-\frac{d\left[N_A\right]}{dt}=k\left[N_A\right]$

Doing integration and rearranging the terms,

$ln\frac{N_{A_o}}{N_A}=kt$

Substituting the values we get $t=30minutes$

So, the minimum time to have spinach before the fight is $30minutes$.