Let the numbers of monosaccharide units be:

• Ribose units $= r$
• 2-deoxyribose units $= d$
• Glucose units $= g$

Since it is a linear octasaccharide, total number of units is $r+d+g=8 \, .$

Use the given mass percentage after complete hydrolysis

On hydrolysis, the constituent monosaccharides are obtained as free sugars.

Given molar masses:

Relate hydrolyzed mass to the original octasaccharide mass

A linear octasaccharide has $8$ monosaccharide units joined by $7$ glycosidic bonds.

Each bond formation removes one $\mathrm{H_2O}$ molecule.

Therefore, $\text{mass of octasaccharide} = \text{sum of monosaccharide masses} - 7\times 18$

Given octasaccharide molar mass $=1024$: $150r+134d+180g - 126 = 1024$ $150r+134d+180g = 1150$

Use the percentage condition

$\frac{134d}{1150}=0.5826$

So, $134d = 1150\times 0.5826 = 669.99 \approx 670$

Thus, $d=\frac{670}{134}=5$

So the number of 2-deoxyribose units is $d=5$

Now use total units equation

$r+g+5=8$ $r+g=3$

Use total hydrolysis mass equation

$150r+134(5)+180g=1150$ $150r+670+180g=1150$ $150r+180g=480$ Divide by $30$: $5r+6g=16$

Also, $r+g=3$

Multiply by $5$: $5r+5g=15$

Subtract from $5r+6g=16$: $g=1$

Therefore, $r=3-1=2$

Final answer

The number of ribose units present in one molecule of octasaccharide is $\boxed{2}$