JEE Advanced · Chemistry · 29. Biomolecules
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): \(\mathrm{H}=1, \mathrm{O}=16\)
- A 1
- B 2
- C 3
- D 8
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\underset{\mathrm{M} . \mathrm{M} .=1024}{\text {Octasaccharide }}+\underset{\mathrm{M} . \mathrm{M} .=126}{7 \mathrm{H}_2 \mathrm{O}} \longrightarrow \text { Ribose }\)\(+\text { 2deoxyribose }+ \text { glucose } \)
\( \text {Total mass }=1024+126=1150\)
\(\begin{aligned}
& 58.26=\frac{134 \times n}{1150} \times 100 \\
& \frac{66.999}{100}=134 n \quad n=4.99=5
\end{aligned}\)
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
\( \text {Total mass }=1024+126=1150\)
\(\begin{aligned}
& 58.26=\frac{134 \times n}{1150} \times 100 \\
& \frac{66.999}{100}=134 n \quad n=4.99=5
\end{aligned}\)
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
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