Question

vii. A living cell contains a solution which
is isotonic with 0.3 M sugar solution.
What osmotic pressure develops when
the cell is placed in 0.1 M KCl solution
at body temperature?
a. 5.08 atm
b. 2.54 atm
c. 4.92 atm
d. 2.46 atm
2. An
sen
i.​

Answer 1

Given:

• Molar concentration of sugar$= 0.3M$

• Molar concentration of$KCL = 1.0M$

To Find:

• Osmotic pressure when  the cell is placed in $KCl$ solution  at body temperature?

Solution:

Quantitative relationship between osmotic pressure and solute concentration, is given by

            ${\displaystyle \Pi =icRT}$

Where,

$\Pi =$ osmotic pressure

$i =$  dimensionless van 't Hoff index

$c =$ the molar concentration of solute

$R =$ the ideal gas constant

$T =$ the temperature in kelvins.

Now,

          $\pi = (i_1 C_1 - i_2 C _2)RT$

          $\pi = (1 \times 0.3 - 2 \times 0.1) 0.0821 \times 300$

          $\pi = 2.463atm$

Thus, The osmotic pressure is  $\pi = 2.463atm$

Answer 2

Explanation:

Molar concentration of sugar= 0.3M=0.3M

Molar concentration ofKCL = 1.0MKCL=1.0M

To Find:

Osmotic pressure when the cell is placed in KClKCl solution at body temperature?

Solution:

Quantitative relationship between osmotic pressure and solute concentration, is given by

{\displaystyle \Pi =icRT}Π=icRT

Where,

\Pi =Π= osmotic pressure

i =i= dimensionless van 't Hoff index

c =c= the molar concentration of solute

R =R= the ideal gas constant

T =T= the temperature in kelvins.

Now,

\pi = (i_1 C_1 - i_2 C _2)RTπ=(i

1

C

1

−i

2

C

2

)RT

\pi = (1 \times 0.3 - 2 \times 0.1) 0.0821 \times 300π=(1×0.3−2×0.1)0.0821×300

\pi = 2.463atmπ=2.463atm

Thus, The osmotic pressure is \pi = 2.463atmπ=2.463atm