The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL$-$1 is _____________.
(Molar mass of glucose = 180 g mol$-$1, R = 0.083 L bar K$-$1 mol$-$1) (Nearest integer)
Answer (integer)
54
Solution
$7.47=\mathrm{C} \times 0.083 \times 300$
<br/><br/>
$(\pi=\mathrm{CRT})$
<br/><br/>
(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)
<br/><br/>
$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$
<br/><br/>
which in $\mathrm{gm} / \mathrm{L}=\frac{7.47}{0.083 \times 300} \times 180$
<br/><br/>
$=54 \mathrm{gm} / \mathrm{l}$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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