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Solutions question

2022 · 24 Jun · Shift 1 · Q14
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Solutions question

2022 · 24 Jun · Shift 1 · Q14

JEE MainChemistrySolutionsNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​. (Molar mass of glucose = 180 g mol −-− 1, R = 0.083 L bar K −-− 1 mol −-− 1) (Nearest integer)
Numerical answer
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Correct answer: 54

  1. Use osmotic pressure formula

For a dilute solution, π=CRT\pi = C R Tπ=CRT where:

  • π=7.47 bar\pi = 7.47\,\text{bar}π=7.47bar
  • R=0.083 L bar K−1 mol−1R = 0.083\,\text{L bar K}^{-1}\text{ mol}^{-1}R=0.083L bar K−1 mol−1
  • T=300 KT = 300\,\text{K}T=300K
  • CCC is molarity in mol L−1\text{mol L}^{-1}mol L−1

So, C=πRT=7.470.083×300C = \frac{\pi}{RT} = \frac{7.47}{0.083 \times 300}C=RTπ​=0.083×3007.47​

  1. Calculate molarity

0.083×300=24.90.083 \times 300 = 24.90.083×300=24.9

C=7.4724.9=0.3 mol L−1C = \frac{7.47}{24.9} = 0.3\,\text{mol L}^{-1}C=24.97.47​=0.3mol L−1

  1. Convert molarity to g L−1^{-1}−1

Mass concentration = molarity ×\times× molar mass

=0.3×180=54 g L−1= 0.3 \times 180 = 54\,\text{g L}^{-1}=0.3×180=54g L−1

  1. Nearest integer

545454

Therefore, the concentration of glucose solution required to be isotonic with blood is: 54 g L−1\boxed{54\,\text{g L}^{-1}}54g L−1​

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