What is the answer to the Biomolecules JEE 2024 polymer U question?
The answer is 93018, obtained by adding the two monomer mass contributions and subtracting the mass of 999 water losses. This Biomolecules JEE 2024 numerical-answer question is from JEE Advanced 2024, Chemistry, Paper 2, Q15, not JEE Main. The supplied record contains no answer options.
Q and R undergo condensation to form polymer U. For one mole of U, the supplied mass contributions and water loss are:
- Mass contributed by Q:
- Mass contributed by R:
- Mass of water eliminated:
Determine the molecular weight of U. These entries are sufficient; you do not need to reconstruct the monomer structures.
The question bank rates this problem medium and sets a 90-second practice target. Neither is an official exam prescription.
How do you calculate the monomer masses and water loss?
By conservation of mass, the polymer mass equals the total reacting monomer mass minus the mass of eliminated water. Every supplied mass corresponds to producing one mole of U. Keep that basis unchanged; these are not masses for a single polymer molecule.
Following the official solution, calculate the two monomer contributions and the water loss separately. Keep the supplied water count visible in the calculation:
- Mass from Q:
- Mass from R:
- Total reacting monomer mass:
- Mass of eliminated water:
Water is subtracted because its mass is no longer retained in U. Mass has not disappeared: it is divided between the polymer and the eliminated water.
Use this explicit mass ledger rather than a repeat-unit shortcut. The required losses are already supplied, so an extra counting assumption is unnecessary.
How does the mass of one mole give the answer 93018?
The subtraction gives a one-mole sample mass of 93018 grams. Dividing by one mole gives the molar mass: the numerical value stays unchanged, but the units change.
The sample mass is measured in grams; the molar mass is measured in grams per mole. They share a numerical value here only because the sample contains exactly one mole.
The requested molecular-weight answer is numerical, so enter:
Check the subtraction physically: the polymer mass must be lower than the initial 111000 grams because water has been eliminated. The calculated mass passes this check.
Why does subtracting 1000 waters give the wrong answer?
Subtracting 1000 waters removes one water too many and produces 93000 instead of 93018. This is an illustrative wrong numerical result, not a documented answer option. The mistake is replacing the supplied elimination count with an assumed “one water per monomer” rule.
The incorrect calculation gives:
The difference is:
The extra subtraction removes one extra water molecule per polymer molecule, or one extra mole of water on the stated one-mole basis. That accounts for the entire error.
The counting principle is conditional. Joining separate units into one open linear chain requires one fewer inter-unit join than the number of units:
If each join eliminates one water molecule, then:
This does not establish unseen structures of Q and R. The official calculation uses the explicitly supplied water-loss mass:
Use the stated loss count. Derive a count from connectivity only when the question gives enough structural information to justify it.
How can you apply this mass balance to peptide questions?
Subtract water when building a peptide; restore that water when working backwards to the free amino acids. The following two questions are original chapter practice, not additional verified JEE PYQs. Both use the same mass balance, with the peptide-bond count made explicit.
What is the molar mass of an open-chain glycine tripeptide?
The tripeptide’s molar mass is 189 grams per mole. Question 1: Three glycine molecules form one open-chain tripeptide. Find its molar mass using:
Three amino-acid residues require two peptide bonds, eliminating two water molecules per tripeptide molecule. Subtract those losses from the combined free-amino-acid mass:
How do you find the free amino acid’s molar mass from a tetrapeptide?
The free amino acid’s molar mass is 89 grams per mole. Question 2: An open-chain tetrapeptide contains four identical amino-acid residues and has the molar mass given below. Find the free amino acid’s molar mass, using the supplied water molar mass:
Three peptide bonds mean three waters were lost. Taking molar masses in grams per mole, write:
Before calculating, choose the direction: building the peptide, subtract water; working backwards to free amino acids, restore water. Then count bonds, not residues.
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Read next: Ray Optics JEE 2022: Two-Lens Matching and Sign Rules.
Frequently asked questions
What is the answer to the Biomolecules JEE 2024 polymer U question?
The numerical answer is 93018. Calculate 500 × 104 + 500 × 118 − 999 × 18 = 93018 grams for one mole of U, giving a molar mass of 93018 grams per mole.
Is the polymer U question from JEE Main or JEE Advanced 2024?
It is from JEE Advanced 2024, Chemistry, Paper 2, Q15, not JEE Main. It is a numerical-answer question, and the supplied record contains no answer options.
Why are 999 waters subtracted instead of 1000 in the polymer U question?
The question explicitly supplies a water-loss mass of 999 × 18 grams for one mole of U. For an open linear chain of 1000 units, 999 joins would eliminate 999 water molecules if each join releases one water molecule. Subtracting 1000 waters instead removes an extra 18 grams on the one-mole basis and gives the incorrect result 93000.
How do you calculate peptide molar mass from amino acid masses?
Add the molar masses of the free amino acids and subtract 18 grams per mole for each peptide bond formed. An open-chain glycine tripeptide has two peptide bonds, so its molar mass is 3 × 75 − 2 × 18 = 189 grams per mole. When working backwards from a peptide to its free amino acids, add back the lost water mass.