A living cell contains a solution which is isotonic with 0.3( M ) sugar solution. What will be the osmotic pressure when the cell is placed in 0.1( M ) KCl solution at body temperature?

Question

A living cell contains a solution which is isotonic with 0.3( M ) sugar solution. What will be the osmotic pressure 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

Tardigrade Admin · Accepted Answer

π =(i1, C1-i2 C2) R T =(1 × 0.3-2 × 0.1) × 0.082 × 310 =2.54 atm

Q. A living cell contains a solution which is isotonic with $0.3 (M)$ sugar solution. What will be the osmotic pressure when the cell is placed in $0.1 (M) K Cl$ solution at body temperature?

5.08 atm

2.54 atm

4.92 atm

2.46 atm

$π = (i_{1}, C_{1} - i_{2} C_{2}) RT$

$= (1 \times 0.3 - 2 \times 0.1) \times 0.082 \times 310$

$= 2.54 a t m$

Q. A living cell contains a solution which is isotonic with 0.3(M) sugar solution. What will be the osmotic pressure when the cell is placed in 0.1(M)KCl solution at body temperature?