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← JEE Main 2022 · 24 Jun · Shift 1
ChemistryInteger+4 / -1Question 83 of 90

JEE Main 2022 · 24 Jun · Shift 1, Chemistry question 83

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\mathrm{gL^{-1}} is ______.

(Molar mass of glucose =180 g mol−1=180\ \mathrm{g\,mol^{-1}}, R=0.083 L bar K−1mol−1R=0.083\ \mathrm{L\,bar\,K^{-1}mol^{-1}}) (Nearest integer)

Answer and worked solution

Official answer: 54

Isotonic. π=CRT\pi=CRT, so C=7.470.083×300=7.4724.9=0.3C=\dfrac{7.47}{0.083\times300}=\dfrac{7.47}{24.9}=0.3 mol L−1^{-1}.

In g/L. 0.3×180=540.3\times180=54 g L−1^{-1}.

Answer: 54