
The osmotic pressure of blood is 8.21 atm at $ {37^0} $ C. How much glucose would be used for an injection that is at the same osmotic pressure as blood?
A) 22.7 $ g{L^{ - 1}} $
B) 58.14 $ g{L^{ - 1}} $
C) 61.26 $ g{L^{ - 1}} $
D) 75.43 $ g{L^{ - 1}} $
Answer
412.5k+ views
Hint :In order to solve this question, you must know the relationship between the osmotic pressure of a solution and the molar concentration of its solute. This relationship was given by the Dutch chemist Jacob van’t Hoff. This equation only holds true for solutions that behave like ideal solutions.
Complete Step By Step Answer:
$ \pi $ = 8.21
V = 1 L
M = 180 (Molar mass of glucose)
T = 273 + 37 = 310 K
R = 0.0821
m = ?
We know that,
$ \pi = CRT $
$ \pi v = {m}{M}RT $
Therefore, $ {m}{v} = {{\pi M}}{{RT}} $
Now put the values, $ {m}{v} $ = $ {{8.21 \times 1 \times 180}}{{0.0821 \times 310}} $
$ {m}{v} $ = 58.06 $ g{L^{ - 1}} $
Hence, option (B) is correct.
Note :
Osmotic pressure can be defined as the minimum pressure that must be applied to the solution to halt the flow of solvent molecules through a semipermeable membrane.
Complete Step By Step Answer:
$ \pi $ = 8.21
V = 1 L
M = 180 (Molar mass of glucose)
T = 273 + 37 = 310 K
R = 0.0821
m = ?
We know that,
$ \pi = CRT $
$ \pi v = {m}{M}RT $
Therefore, $ {m}{v} = {{\pi M}}{{RT}} $
Now put the values, $ {m}{v} $ = $ {{8.21 \times 1 \times 180}}{{0.0821 \times 310}} $
$ {m}{v} $ = 58.06 $ g{L^{ - 1}} $
Hence, option (B) is correct.
Note :
Osmotic pressure can be defined as the minimum pressure that must be applied to the solution to halt the flow of solvent molecules through a semipermeable membrane.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
