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JEE Advanced Chemistry Biomolecules 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Chemistry Question (2025) — Solution

Question

A linear octasaccharide (molar mass =1024 ~g ~mol ^ -1 ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 \%( w / 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 Use : Molar mass ( . in . g mol ^ -1 ) : ribose =150,2-deoxyribose =134, glucose =180; Atomic mass (in amu): H =1, O =16

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

M . M .=1024 Octasaccharide + M . M .=126 7 H _2 O aligned & Ribose + 2deoxyribose + glucose \\ & Total mass =1024+126=1150 aligned aligned & 58.26= 134 n 1150 100 \\ & 66.999 100 =134 n n=4.99=5 aligned 5 units of 2-Deoxyribose 1150=(5 150)+( x 150)+( y 180) aligned & 1150= 750 _ 5 unit + 150 x _ 300 + 180 y _ 180 \\ & 2 unit 1 unit aligned n = 2.00

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