The osmotic pressure of blood is 7.65 atm. at 310 K. An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume)

Question

The osmotic pressure of blood is 7.65 atm. at 310 K. An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume) (A) 5.41 (B) 3.54 (C) 4.53 (D) 53.4

Tardigrade Admin · Accepted Answer

Since glucose and blood are isotonic, the osmotic pressures are the same. Therefore,

π_{glucose} = π_{blood}

Or

7.65 \times V = \frac{w}{180} \times 0.0821 \times 310

[w = amount of glucose present in V litres of solution]
Or

7.65 \times 180 = \frac{w}{V} \times 0.0821 \times 310

Or

\frac{w}{V} = \frac{7.65 \times 180}{0.0821 \times 310}

= 54.1 g L^{- 1}

or

5.41%

Q. The osmotic pressure of blood is 7.65 atm. at 310 K. An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume)

$5.41$

$3.54$

$4.53$

$53.4$

Q. The osmotic pressure of blood is 7.65 atm. at 310 K. An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume)

5.41

3.54

4.53

53.4

Since glucose and blood are isotonic, the osmotic pressures are the same. Therefore, πglucose​=πblood​ Or 7.65×V=180w​×0.0821×310 [w = amount of glucose present in V litres of solution] Or 7.65×180=Vw​×0.0821×310 Or Vw​=0.0821×3107.65×180​ =54.1gL−1 or 5.41%

$5.41$

$3.54$

$4.53$

$53.4$