Question 16

NUMERICALMEDIUM

A linear octasaccharide (molar mass =1024 g mol−1= 1024\text{ 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)58.26\% \text{ (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\text{g mol}^{-1}): ribose =150= 150, 2-deoxyribose =134= 134, glucose =180= 180; Atomic mass (in amu): H=1,O=16H = 1, O = 16

Correct Answer: 2

Detailed Solution

  1. Hydrolysis of octasaccharide (8 units) involves 7 water molecules: Octasaccharide+7H2O→Monosaccharides\text{Octasaccharide} + 7H_2O \rightarrow \text{Monosaccharides} Total mass of products =1024+7(18)=1024+126=1150 g/mol= 1024 + 7(18) = 1024 + 126 = 1150\text{ g/mol}.

  2. Mass of 2-deoxyribose formed =58.26% of 1150=0.5826×1150=669.99≈670 g= 58.26\% \text{ of } 1150 = 0.5826 \times 1150 = 669.99 \approx 670\text{ g}. Number of 2-deoxyribose units =670134=5= \frac{670}{134} = 5.

  3. Total units in octasaccharide =8= 8. Remaining units =8−5=3= 8 - 5 = 3 (Ribose + Glucose). Mass of these 3 units =1150−670=480 g= 1150 - 670 = 480\text{ g}. Let xx be the number of ribose units and yy be the number of glucose units. x+y=3x + y = 3 150x+180y=480150x + 180y = 480 Divide by 30: 5x+6y=165x + 6y = 16 Substitute y=3−xy = 3-x: 5x+6(3−x)=165x + 6(3 - x) = 16 5x+18−6x=16  ⟹  −x=−2  ⟹  x=25x + 18 - 6x = 16 \implies -x = -2 \implies x = 2. Number of ribose units is 2.

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