JEE Advanced 2025 Paper 2 · Q16 · Mole Concept & Stoichiometry
A linear octasaccharide (molar mass = 1024 g/mol) 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 (g/mol): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass: $H = 1$, $O = 16$.
Reveal answer + step-by-step solution
Correct answer:2
Solution
Complete hydrolysis of a linear octasaccharide breaks 7 glycosidic linkages, adding 7 $H_2O$ ($7 \times 18 = 126$ g per mole of octasaccharide). Total mass of monosaccharides per mole = $1024 + 126 = 1150$ g. Mass of 2-deoxyribose = $0.5826 \times 1150 = 670$ g; moles of 2-deoxyribose = $670/134 = 5$. Remaining $8 - 5 = 3$ monosaccharides are split between ribose and glucose with combined mass $1150 - 670 = 480$ g. Let $r$ = number of ribose units, then glucose units = $3 - r$. $150 r + 180 (3 - r) = 480 \to 150 r + 540 - 180 r = 480 \to -30 r = -60 \to r = 2$.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →