A linear octasaccharide (molar mass ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 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 ): ribose = 150, 2-deoxyribose = 134, glucose = 180;
Atomic mass (in amu):
Use : Molar mass (in ): ribose = 150, 2-deoxyribose = 134, glucose = 180;
Atomic mass (in amu):
Answer: 2
Solution
Step 1: Calculate total mass after hydrolysis:
Octasaccharide + 7H₂O → 8 monosaccharides
Mass of water added = 7 × 18 = 126 g/mol
Total mass = 1024 + 126 = 1150 g/mol
Step 2: Find number of 2-deoxyribose units:
58.26% of 1150 = mass of 2-deoxyribose
So there are 5 units of 2-deoxyribose.
Step 3: Find ribose and glucose units:
Let x = ribose units, y = glucose units
Total units: x + y + 5 = 8 → x + y = 3
Total mass: 150x + 180y + (5 × 134) = 1150
150x + 180y + 670 = 1150
150x + 180y = 480
Dividing by 30: 5x + 6y = 16
From x + y = 3: y = 3 - x
Substituting: 5x + 6(3 - x) = 16
5x + 18 - 6x = 16
-x = -2
x = 2
Therefore, there are 2 ribose units.
Octasaccharide + 7H₂O → 8 monosaccharides
Mass of water added = 7 × 18 = 126 g/mol
Total mass = 1024 + 126 = 1150 g/mol
Step 2: Find number of 2-deoxyribose units:
58.26% of 1150 = mass of 2-deoxyribose
So there are 5 units of 2-deoxyribose.
Step 3: Find ribose and glucose units:
Let x = ribose units, y = glucose units
Total units: x + y + 5 = 8 → x + y = 3
Total mass: 150x + 180y + (5 × 134) = 1150
150x + 180y + 670 = 1150
150x + 180y = 480
Dividing by 30: 5x + 6y = 16
From x + y = 3: y = 3 - x
Substituting: 5x + 6(3 - x) = 16
5x + 18 - 6x = 16
-x = -2
x = 2
Therefore, there are 2 ribose units.
JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer