Question 16

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

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

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