Living cell =0.9gm in 100 gm of solution %w/w=0.9 Solution is have equal moles of glucose and water =0.5 Weight of solution =0.5×180+0.5×18=99gm %w/w≃90% Concentrated solution = Cell will shrink.
Assume a living cell with 0.9 %(ω / ω) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose…
JEE Main 2025 — Chemistry Physical Chemistry
Assume a living cell with 0.9%(ω/ω) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
show no change in volume since solution is 0.9%(ω/ω)
- B
shrink since solution is 0.5%(ω/ω)
- C
swell up since solution is 1%(ω/ω)
- D
None of these
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Physical Chemistry in other years
More JEE Main Physical Chemistry questions
What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?
  Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?
At 298 K, the mole percentage of $\mathrm{N}_{2}(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\mathrm{N}_{2}(\mathrm{~g})$ in water at 298 K ? ($\mathrm{K}_{\mathrm{H}}$ for $\mathrm{N}_{2}$ is $6.5 \times 10^{7} \mathrm{~mm} \mathrm{Hg}$)
Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL. (Nearest integer) [Given: $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text {red }}^{\mathrm{o}}=+0.34 \mathrm{~V}$ $\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text {red }}^{\mathrm{o}}=+1.23 \mathrm{~V}$ Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$ Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$ Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$]
Arrange the following resultant mixtures in increasing order of their pH values A. $10$ mL $0.2$ M Ca(OH)$_2$ + $25$ mL $0.1$ M HCl B. $10$ mL $0.01$ M H$_2$SO$_4$ + $10$ mL $0.01$ M Ca(OH)$_2$ C. $10$ mL $0.1$ M H$_2$SO$_4$ + $10$ mL $0.1$ M KOH Choose the correct answer from the options given below:
Other Chemistry topics for JEE Main
Keep practicing Physical Chemistry
Work through every JEE Main Physical Chemistry PYQ, year by year.