Question

The circulation of blood in human body supplies $O_2$ and releases $C{O}_2$. The concentration of $O_2$ and $C{O}_2$ is variable but on the average, $100mL$ blood contains $0.02g$ of $O_2$ and $0.08gC{O}_2$. The volume of $O_2$ and $C{O}_2$ at $1atm$ and body temperature ${37}^{\circ }C$, assuming $10L$ blood in human body is:

• A. $2L,4L$
• B. $1.5L,4.5L$
• C. $1.59L,4.62L$
• D. $3.82L,2.64L$

Answer 1

The correct option is A $2L,4L$

Given,

$100ml$ blood contains $0.02g$ $O_2$ & $0.08g$ $C{O}_2$

$\therefore$ No. of moles of $O_2$ in $100ml$ blood=$\frac{0.02g}{32g/mol}=0.000625\approx 0.001$ moles

Also, No. of moles of $C{O}_2$ in $100ml$ blood=$\frac{0.08g}{44g/mol}=0.00181\approx 0.002$ moles

Now, we know $1$ mole $O_2=22.4L$ $O_2$

$\Rightarrow 0.001$ mole $O_2=22.4\times {10}^{-3}L$ $O_2$

Also, $1$ mole $C{O}_2=22.4L$ $C{O}_2$

$0.002$ mole $C{O}_2=22.4\times 0.002$

$=44.8\times {10}^{-3}L$ of $C{O}_2$

$\Rightarrow 100ml$ blood has $422.4\times {10}^{-3}L$ $O_2$

$\therefore 10L$ blood has $\frac{22.4\times {10}^{-3}\times 10}{100\times {10}^{-3}}=2.24L$ of $O_2$

Also, $100ml$ blood has $44.8\times {10}^{-3}L$ $C{O}_2$

$\therefore 10L$ blood has $\frac{44.8\times {10}^{-3}\times 10}{100\times {10}^{-3}}=4.48L$ of $C{O}_2$

$\Rightarrow 10L$ blood has $2L$ $O_2$ & $4L$ $C{O}_2$.