Question

Solve question no. 8 and 9. Question is in attachment. ​

Answer 1

Question :-9

$\bigstar \: \pink{ \bf \underline{Given \: information :-}}$

• Human lungs can absorb 8gm $O_2$ per hour by respiration
• All oxygen atoms are converted into $C_6H_{12}O_6$

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$\bigstar \: \blue{ \bf{ \underline{To \: Find :-}}}$

• Time taken to produce 180 gm $C_6H_{12}O_6$

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$\bigstar \: \green { \bf{ \underline{ Concept :-}}}$

In this question we have been given the absorption rate of O2, and we have to find the time taken to produce 180 gm of $C_6H_{12}O_6$. So for that, first of all we will calculate the mass of O2 present in this glucose. And then we will divide it's mass by the mass of O2 absorbed per hour to obtain the time taken. So let's start. :D

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$\bigstar \: \purple{ \bf{ \underline{Solution :-}}}$

$\sf \longrightarrow \: Mass \: of \: O_2 = 180 \times \dfrac{16 \times 6}{180} \\ \\ \sf \longrightarrow \: Mass \: of \: O_2 =16 \times 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \longrightarrow \: Mass \: of \: O_2 = \underline{\boxed{ \sf96 \: gm}}. \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the mass of O2 in 180 gm glucose is 96 gm

Now we will find out the time taken, thus, we have,

$\sf \longrightarrow Time \: taken = \dfrac{96}{8} \\ \\ \sf \longrightarrow Time \: taken =12 \:$

Therefore, the time taken to produce 180 gm of glucose will be 12 hours i.e. option (b)

Answer 2

Answer:

I hope it helped answer in the attachment

pls move the attachment to the right