A linear octasaccharide (molar mass = 1024 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) 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 ): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu) : H = 1, O = 16

To determine the number of ribose units present in one molecule of the linear octasaccharide, we can analyze the components and the stoichiometry of the complete hydrolysis of the octasaccharide.

First, let's understand the hydrolysis reaction of a linear octasaccharide.
An octasaccharide consists of 8 monosaccharide units linked together by glycosidic bonds. In a linear octasaccharide, there are 7 glycosidic bonds linking the 8 monosaccharide units.
When it undergoes complete hydrolysis, it reacts with water to break these 7 bonds. Thus, 7 molecules of water ($H_{2}O$) are consumed for every molecule of octasaccharide hydrolyzed:
$\text{Octasaccharide}+7H_{2}O\to \text{8 Monosaccharides}$

Let's calculate the total mass of the monosaccharides produced from one mole of the octasaccharide (molar mass = 1024 g mol^-1):
Molar mass of $H_{2}O$ = $2\times 1+16=18\text{g}{\text{mol}}^{-1}$
Mass of 7 moles of water = $7\times 18=126\text{g}$
By conservation of mass, the total mass of the mixture of monosaccharides produced is:
$\text{Total mass of products}=1024+126=1150\text{g}$

The problem states that the amount of 2-deoxyribose formed is 58.26% (w/w) of the total mass of the monosaccharides produced.
Let's find the mass of 2-deoxyribose in the mixture:
$\text{Mass of 2-deoxyribose}=1150\times \frac{58.26}{100}=669.99\approx 670\text{g}$

Now, let's find the number of moles of 2-deoxyribose units in one mole of the octasaccharide.
Molar mass of 2-deoxyribose = 134 g mol^-1
$\text{Number of 2-deoxyribose units}=\frac{670}{134}=5$

Let $x$ be the number of ribose units and $y$ be the number of glucose units in the octasaccharide.
Since the octasaccharide is composed of 8 monosaccharide units (ribose, 2-deoxyribose, and glucose), we have:
$x+5+y=8$
This simplifies to:
$x+y=3$

Now, we use the remaining mass of the monosaccharide products to find $x$ and $y$.
The combined mass of ribose and glucose units produced is:
$\text{Mass of (ribose + glucose)}=1150-670=480\text{g}$

Given the molar masses of ribose (150 g mol^-1) and glucose (180 g mol^-1), we can set up the mass equation:
$150x+180y=480$

We can solve this system of linear equations:
1) $x+y=3\Rightarrow y=3-x$
Substitute this into the mass equation:
$150x+180(3-x)=480$
$150x+540-180x=480$
$-30x=-60$
$x=2$

Therefore, the number of ribose units present in one molecule of the octasaccharide is 2.