Biomolecules JEE 2025: Octasaccharide Question Solved

What is the Biomolecules JEE 2025 octasaccharide question?

The percentage denominator is 1150 g of monosaccharides, not 1024 g of parent carbohydrate. This Biomolecules JEE 2025 question appeared in JEE Advanced 2025, Paper 2, Chemistry, as a numerical-answer question. It is not a JEE Main question.

An unbranched carbohydrate contains eight monosaccharide units. Its molar mass is:

M_{octasaccharide} = 1024 g m o l^{- 1}

Complete hydrolysis produces ribose, 2-deoxyribose and glucose. 2-deoxyribose contributes 58.26% by mass of the total monosaccharide products. Find the number of ribose units in one original octasaccharide molecule.

Use these supplied values:

Ribose molar mass: $150 g m o l^{- 1}$
2-deoxyribose molar mass: $134 g m o l^{- 1}$
Glucose molar mass: $180 g m o l^{- 1}$
Atomic masses:

H = 1 a m u, O = 16 a m u

The question bank tags this as medium difficulty, with 120 seconds as a suggested solving time. These are question-bank guidance, not an official difficulty classification or a guaranteed completion time.

Why must water be added before applying the percentage?

Hydrolysis incorporates water into the monosaccharides, so their combined mass exceeds the starting carbohydrate mass. Following the official solution, choose one mole of octasaccharide, initially weighing 1024 g.

Eight units in a linear chain have seven glycosidic bonds. Complete hydrolysis consumes one water molecule per bond: seven water molecules per parent molecule, or seven moles of water per mole of parent carbohydrate. $N_{bonds} = 8 - 1 = 7$

M (H_{2} O) = 2 \times 1 + 16 = 18 g m o l^{- 1}

m_{water incorporated} = 7 \times 18 = 126 g

Add this incorporated water to the parent mass to obtain the total monosaccharide mass. On the one-mole basis:

m_{monosaccharides} = 1024 + 126 = 1150 g

This is the combined mass of the monosaccharides, not the mass of the aqueous solution. Count the water chemically incorporated during bond cleavage, but exclude excess solvent water from the percentage denominator.

How does 58.26% give five 2-deoxyribose units?

Applying the percentage to 1150 g gives approximately 670 g of 2-deoxyribose. Dividing by its supplied molar mass gives five moles per mole of parent carbohydrate, hence five units per original molecule. The small decimal discrepancy comes from the reported percentage being rounded.

Mass percentage = \frac{m_{2-deoxyribose}}{m_{total monosaccharides}} \times 100

m_{2-deoxyribose} = \frac{58.26}{100} \times 1150 = 669.99 g \approx 670 g

Check that 670 g reproduces the stated percentage. Its percentage rounds to the supplied 58.26%:

\frac{670}{1150} \times 100 = 58.260869 \dots % \approx 58.26 %

Then divide by the molar mass:

n_{2-deoxyribose} = \frac{670 g}{134 g m o l^{- 1}} = 5 m o l

This rounding is justified by the percentage check. It is not permission to round any fractional unit count to the nearest integer.

How do we obtain two ribose units and verify 02.00?

The remaining three units must be two ribose units and one glucose unit. Use both the remaining unit count and remaining product mass to establish that result. A unit count alone cannot distinguish ribose from glucose.

After accounting for five 2-deoxyribose units, the remaining count is: $8 - 5 = 3$

Define the unknown counts and write their sum:

r = number of ribose units, g = number of glucose units

$r + g = 3$ The remaining product mass is 480 g on our one-mole basis. Using the supplied molar masses gives the second equation: $1150 - 670 = 480 g$ $150 r + 180 g = 480$

Substitute for the glucose count and simplify:

g & = 3 - r 150 r + 180 (3 - r) & = 480 150 r + 540 - 180 r & = 480 540 - 30 r & = 480 30 r & = 60 r & = 2 g & = 3 - 2 = 1

The complete composition is two ribose, five 2-deoxyribose and one glucose unit. Check the total number of units, then reconstruct the original molar mass by subtracting the water lost when seven bonds formed: $2 + 5 + 1 = 8$

M_{parent} = 2 \times 150 + 5 \times 134 + 1 \times 180 - 7 \times 18 = 1024 g m o l^{- 1}

The unit count, parent molar mass and product percentage all agree. The requested answer is two ribose units, matching the supplied answer key:

02.00

Why is taking 58.26% of 1024 incorrect?

That calculation applies a product percentage to the parent carbohydrate’s mass. The 1024 g describes the material before hydrolysis, while 58.26% describes the combined monosaccharides after hydrolysis. This numerical-answer question has no supplied multiple-choice options; the calculation below is an incorrect route, not an answer option.

The faulty mass calculation and its implied unit count are:

m_{2-deoxyribose, wrong} = 0.5826 \times 1024 = 596.5824 g

\frac{596.5824}{134} \approx 4.452

That is inconsistent with an integer composition. Compare it with the count obtained from the correct product mass:

\frac{669.99}{134} \approx 4.99993

The count 4.452 comes from a mass-basis error, whereas 4.99993 reflects the stated percentage’s rounding. Rounding the wrong result cannot repair the chemistry.

Use this repair rule:

Identify the material named in the percentage denominator.
Include chemically consumed water in the product mass.
Only then apply the percentage and convert product mass to a unit count.

How can you test the method on two different carbohydrates?

Work backwards from the free sugars: add their masses, then subtract the water lost during chain formation. These two questions are original practice, not additional JEE PYQs. Their hypothetical compositions are exercise inputs.

Original practice 1: A linear trisaccharide contains two ribose units and one glucose unit. Using the molar masses below, calculate its molar mass and the mass of monosaccharides obtained by completely hydrolysing one mole.

M_{ribose} = 150 g m o l^{- 1}, M_{glucose} = 180 g m o l^{- 1}

Three units give two bonds. Subtract the mass of two moles of water from the free-sugar mass:

m_{free sugars} = 2 \times 150 + 180 = 480 g

M_{parent} = 480 - 2 \times 18 = 444 g m o l^{- 1}

Hydrolysis adds back 36 g of water lost during chain formation. This yields 480 g of monosaccharides:

m_{water added} = 2 \times 18 = 36 g

m_{products} = 444 + 36 = 480 g

Original practice 2: A linear octasaccharide consists entirely of glucose units. Using the supplied glucose molar mass, calculate its molar mass and the mass increase when one mole undergoes complete hydrolysis.

M_{glucose} = 180 g m o l^{- 1}

Eight units give seven bonds. Subtract seven moles of water to obtain the parent molar mass; complete hydrolysis adds that water back:

m_{free glucose} = 8 \times 180 = 1440 g

M_{parent} = 1440 - 7 \times 18 = 1314 g m o l^{- 1}

$Δ m = 7 \times 18 = 126 g$ Before applying any product percentage, account for the water consumed. Complete hydrolysis of a linear chain adds one fewer water molecule than its number of monosaccharide units. Per parent molecule:

n monosaccharide units \Rightarrow (n - 1) water molecules consumed

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

If that step was the hard part, work through Matrices and Determinants JEE 2010: Worked Solution.

Frequently asked questions

Was the Biomolecules 2025 octasaccharide question asked in JEE Main or Advanced?

The question appeared in JEE Advanced 2025, Paper 2, Chemistry, as a numerical-answer question. It was not a JEE Main question.

Why is the percentage denominator 1150 g and not 1024 g?

The 58.26% refers to the total monosaccharide products after hydrolysis, not the parent carbohydrate. The linear octasaccharide has seven glycosidic bonds, so hydrolysing one mole consumes 7 × 18 = 126 g of water. The product mass is therefore 1024 + 126 = 1150 g, excluding excess solvent water.

How does 58.26% give five 2-deoxyribose units?

On a one-mole basis for the parent carbohydrate, the 2-deoxyribose mass is 0.5826 × 1150 = 669.99 g, approximately 670 g. Dividing by its molar mass gives 670/134 = 5 moles, corresponding to five units per parent molecule. This is consistent with the reported percentage because (670/1150) × 100 rounds to 58.26%.

How many ribose units are in the JEE Advanced 2025 octasaccharide?

The octasaccharide contains two ribose units, matching the answer key value 02.00. After accounting for five 2-deoxyribose units, the remaining three units contribute 480 g of products per mole of parent carbohydrate. Solving r + g = 3 and 150r + 180g = 480 gives r = 2 ribose units and g = 1 glucose unit.

Biomolecules JEE 2025: Octasaccharide Question Solved

What is the Biomolecules JEE 2025 octasaccharide question?

Why must water be added before applying the percentage?

How does 58.26% give five 2-deoxyribose units?

How do we obtain two ribose units and verify 02.00?

Why is taking 58.26% of 1024 incorrect?

How can you test the method on two different carbohydrates?

Frequently asked questions

Was the Biomolecules 2025 octasaccharide question asked in JEE Main or Advanced?

Why is the percentage denominator 1150 g and not 1024 g?

How does 58.26% give five 2-deoxyribose units?

How many ribose units are in the JEE Advanced 2025 octasaccharide?

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