Q16 / 48 · JEE Advanced 2025 (Paper 2) · Chemistry
A linear octasaccharide (molar mass $=1024 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is $58.26 \%(\mathrm{w} / \mathrm{w})$ 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 $\qquad$
Use : Molar mass $\left(\right.$ in $\left.\mathrm{g} \mathrm{mol}^{-1}\right)$ : ribose $=150,2$-deoxyribose $=134$, glucose $=180$; Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$
Correct answer: 2
Solution
$\underset{\mathrm{M} . \mathrm{M} .=1024}{\text { Octasaccharide }}+\underset{\mathrm{M} . \mathrm{M} .=126}{7 \mathrm{H}_2 \mathrm{O}} \longrightarrow \begin{aligned} & \text { Ribose }+ \text { 2deoxyribose }+ \text { glucose } \\ & \text { Total mass }=1024+126=1150\end{aligned}$
$\begin{aligned} & 58.26=\frac{134 \times n}{1150} \times 100 \\ & \frac{66.999}{100}=134 n \quad n=4.99=5 \end{aligned}$
5 units of 2-Deoxyribose $1150=(5 \times 150)+(\mathrm{x} \times 150)+(\mathrm{y} \times 180)$
$\begin{aligned} & 1150=\underbrace{750}_{5 \text { unit }}+\underbrace{150 x}_{300}+\underbrace{180 y}_{180} \\ & \qquad \quad \qquad \quad \text{2 unit} \quad \text{1 unit}\end{aligned}$ n = 2.00
$\begin{aligned} & 58.26=\frac{134 \times n}{1150} \times 100 \\ & \frac{66.999}{100}=134 n \quad n=4.99=5 \end{aligned}$
5 units of 2-Deoxyribose $1150=(5 \times 150)+(\mathrm{x} \times 150)+(\mathrm{y} \times 180)$
$\begin{aligned} & 1150=\underbrace{750}_{5 \text { unit }}+\underbrace{150 x}_{300}+\underbrace{180 y}_{180} \\ & \qquad \quad \qquad \quad \text{2 unit} \quad \text{1 unit}\end{aligned}$ n = 2.00
Hence, the correct answer is 2.