Question

Osmotic pressure of a blood sample is 4.92 atm
at 27°C. Which of the following is not isotonic with
blood sample?
(1) 3.6% (m/m%) glucose aqueous solution
(2) 1.2% (m/m%) urea aqueous solution
(3) 0.585% (m/m%) NaCl aqueous solution
(4) 1.7% (m/m%) NaNo, aqueoous solution
​

Answer 1

Given:

The osmotic pressure of blood sample = 4.92 atm

temperature = 27°C = 300K

To Find:

solution isotonic with the blood sample

Solution:

Now, we know that the expression for osmotic pressure is

osmotic pressure ( π ) = iCRT

Now, for an isotonic solution, the concentration of both the sample would be same.

For the concentration of blood sample

4.92 = 1 × C × 0.082 × 300

C = 0.2

For the concentration of glucose aqueous solution

C = $\dfrac{3.6 \times 1000}{100 \times molar mass}$

the molar mass of glucose = 180

C = $\dfrac{3.6 \times 1000}{100 \times 180}$ = 0.2

thus isotonic since, concentration of blood sample = conc. of glucose aqueous solution.

For the concentration of urea aqueous solution

C = $\dfrac{1.2 \times 1000}{100 \times molar mass}$

the molar mass of urea = 60

C = $\dfrac{1.2 \times 1000}{100 \times 60}$ = 0.2

thus isotonic since, concentration of blood sample = conc. of urea aqueous solution

For the concentration of NaCl aqueous solution

C = $\dfrac{0.585\times 2 \times 1000}{100 \times molar mass}$

the molar mass of NaCl  = 58

C = $\dfrac{0.585\times 2 \times 1000}{100 \times 58}$ = 0.2

thus isotonic since, concentration of blood sample = conc. of NaCl aqueous solution

For the concentration of NaNo3 aqueous solution

C = $\dfrac{1.7\times 2 \times 1000}{100 \times molar mass}$

the molar mass of NanO3  = 85

C = $\dfrac{1.7\times 2 \times 1000}{100 \times 85}$ = 0.4

thus not isotonic since, concentration of blood sample ≠ conc. of NaNo3 aqueous solution

Hence the answer is (4)