Chemistry · Biomolecules

JEE Advanced 2025 — Paper 2 — Question 45

A linear octasaccharide (molar mass $=1024 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose

formed is $58.26 \%(\mathrm{w} / \mathrm{w})$ of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is $\qquad$ . Use : Molar mass (in g $\mathrm{mol}^{-1}$ ): ribose $=150,2$ -deoxyribose $=134$ , glucose $=180$ ;

Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\underset{\text { M.M. }=1024}{\text { Octasacharide }}+\underset{\text { M.M. }=126}{7 \mathrm{H}_{2} \mathrm{O}} \longrightarrow$ Ribose + 2deoxyribose + glucose

Total mass $=1024+126=1150$

$58.26=\frac{134 \times n}{1150} \times 100$

$\frac{66.999}{100}=134 n \quad n=4.99=5$

5 units of 2-Deoxyribose $1150=(5 \times 150)+(\mathrm{x} \times 150)+(\mathrm{y} \times 180)$

$1150=\underbrace{750}_{5 \text { unit }}+\underbrace{1500}_{2 \text { unit }}+\underbrace{180}_{1 \text { unit }}$

$\mathrm{n}=2.00$

Answer key and solution verified before publishing.

Practise Biomolecules

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Chemistry
Chapter: Biomolecules
Topic: Introduction to Biomolecules & Carbohydrates