You visited us 1 times! Enjoying our articles? Unlock Full Access!

Diffusion

In the depict...

Question

In the depicted figure, the concentration of $C O_{2}$ in the cell is (i) whereas the concentration of $O_{2}$ in the cell is (ii). Here, (i) and (ii), respectively are

More and more

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Less and more

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

More and less

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Less and less

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Less and more
According to osmosis, substances move from their higher concentration to their lower concentration by the process of diffusion through a semi-permeable plasma membrane. In the given figure, $C O_{2}$ is diffusing inside the cell which means that $C O_{2}$ inside the cell is less, while $O_{2}$ is diffusing out of the cell which means that $O_{2}$ is more inside the cell.

Suggest Corrections

Similar questions

0.4 M

0.1 M

0.4 M

2 M

3 M

C o^{2 +}

\begin{matrix} P t H_{2} P_{1} \end{matrix} | \begin{matrix} H C l \end{matrix} | \begin{matrix} H_{2} P t P_{2} \end{matrix}

P_{1} > P_{2}

\begin{matrix} P t H_{2} \end{matrix} ∣ ∣ \begin{matrix} H C l C_{1} \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} H C l C_{2} \end{matrix} ∣ ∣ \begin{matrix} H_{2} P t \end{matrix}

C_{2} > C_{1}

E_{c e l l} = - \frac{0.059}{2} l o g \frac{P_{2}}{P_{1}}

E_{c e l l} = - \frac{0.059}{1} l o g \frac{C_{1}}{C_{2}}

Join BYJU'S Learning Program