IIT JAM Exam  >  IIT JAM Questions  >  The number of water molecules present in a 30... Start Learning for Free
The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadro’s number = 6 × 1023.
  • a)
    224
  • b)
    9
  • c)
    140
  • d)
    28
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The number of water molecules present in a 300 residue soluble protein...
To solve this problem, we need to calculate the volume of the protein and the volume of water present in it.

1. Volume of the protein:
Given that the diameter of the protein is 2 nm, we can calculate the radius as 1 nm (since radius = diameter/2). Using the formula for the volume of a sphere (V = 4/3 * π * r^3), we can calculate the volume of the protein.

V_protein = 4/3 * π * (1 nm)^3

2. Volume of water:
We are given that the protein has 20% (V/V) water. This means that 20% of the total volume of the protein is occupied by water. We can calculate the volume of water using this percentage.

V_water = 20% * V_protein

3. Number of water molecules:
To calculate the number of water molecules, we need to know the density of water and the molar mass of water. Given that the density of water is 1000 kg/m^3 and Avogadro's number is 6 * 10^23, we can calculate the molar mass of water using its chemical formula (H2O) and the atomic masses of hydrogen and oxygen.

Molar mass of water = (2 * atomic mass of hydrogen) + atomic mass of oxygen

Now, we can use the formula to calculate the number of water molecules:

Number of water molecules = (V_water * density of water) / molar mass of water

4. Calculation:
Let's plug in the values and calculate the number of water molecules:

V_protein = 4/3 * π * (1 nm)^3 = 4/3 * 3.14 * (1 nm)^3
V_water = 20% * V_protein
Number of water molecules = (V_water * density of water) / molar mass of water

Substituting the given values:
V_protein = 4/3 * 3.14 * (1 nm)^3 = 4/3 * 3.14 * (10^-9 m)^3
V_water = 20% * V_protein
Number of water molecules = (V_water * density of water) / molar mass of water

After calculating these values, we find that the number of water molecules is approximately 28. Therefore, the correct answer is option D.
Explore Courses for IIT JAM exam
The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer?
Question Description
The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer?.
Solutions for The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The number of water molecules present in a 300 residue soluble protein of spherical shape (diameter = 2 nm) having 20% (V/V) water is closed to Assume: density of water = 1000 kg/m3, Avogadros number = 6 1023.a)224b)9c)140d)28Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
Explore Courses for IIT JAM exam

Suggested Free Tests

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev