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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Chemistry
Chapter
Biomolecules

JEE Advanced 2025 · Paper 2 · Chemistry · Biomolecules

A linear octasaccharide (molar mass =1024 ~g ~mol^ -1 ) on complete...

A linear octasaccharide (molar mass =1024 g mol−1=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%(w/w)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 (\left(\right. in gmol−1)\left.\mathrm{g} \mathrm{mol}^{-1}\right) : ribose =150,2=150,2-deoxyribose =134=134, glucose =180=180; Atomic mass (in amu): H=1,O=16\mathrm{H}=1, \mathrm{O}=16

Correct answer

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Step-by-step solution

 Octasaccharide M.M.=1024+7H2OM.M.=126⟶\underset{\mathrm{M} . \mathrm{M} .=1024}{\text{ Octasaccharide }}+\underset{\mathrm{M} . \mathrm{M} .=126}{7 \mathrm{H}_2 \mathrm{O}} \longrightarrow Ribose + 2deoxyribose + glucose  Total mass =1024+126=1150\begin{aligned} & \text{ Ribose }+ \text{ 2deoxyribose }+ \text{ glucose } \\ & \text{ Total mass }=1024+126=1150\end{aligned}58.26=134×n1150×10066.999100=134nn=4.99=5\begin{aligned} & 58.26=\frac{134 \times n}{1150} \times 100 \\ & \frac{66.999}{100}=134 n \quad n=4.99=5 \end{aligned}5 units of 2-Deoxyribose $1150=(5 \times 150)+(\mathrm{x} \times 150)+(\mathrm{y} \times 180)\begin{aligned} & 1150=\underbrace{750}_{5 \text{ unit }}+\underbrace{150 x}_{300}+\underbrace{180 y}_{180} \\ & \qquad \quad \qquad \quad \text{2 unit} \quad \text{1 unit}\end{aligned}$$$ n = 2.00
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