JEE MAIN - Chemistry (2022 - 24th June Morning Shift - No. 14)
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
54
Explanation
$7.47=\mathrm{C} \times 0.083 \times 300$
$(\pi=\mathrm{CRT})$
(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)
$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$
which in $\mathrm{gm} / \mathrm{L}=\frac{7.47}{0.083 \times 300} \times 180$
$$ =54 \mathrm{gm} / \mathrm{l} $$
$(\pi=\mathrm{CRT})$
(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)
$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$
which in $\mathrm{gm} / \mathrm{L}=\frac{7.47}{0.083 \times 300} \times 180$
$$ =54 \mathrm{gm} / \mathrm{l} $$
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