One gram of a polysaccharide composed of 1000 glucose units has the same effect on osmolarity as that of
1 mg glucose
Polysaccharide Osmolarity Effect
The question asks about the osmolarity effect of one gram of a polysaccharide compared to different amounts of glucose. Osmolarity is a measure of the concentration of solute particles in a solution. It depends on the number of moles of solute, not the mass or size of the solute molecules (for ideal solutions).
Glucose Units in Polysaccharide
The polysaccharide is composed of 1000 glucose units. When glucose units link together to form a polysaccharide chain, water molecules are removed in the process. If 1000 glucose units link to form a linear polysaccharide, there will be 999 glycosidic bonds formed, and 999 water molecules will be released.
- Molar mass of glucose (\(C_6H_{12}O_6\)) is approximately \(180 \, \text{g/mol}\).
- Molar mass of water (\(H_2O\)) is approximately \(18 \, \text{g/mol}\).
- The molar mass of a polysaccharide made of 1000 glucose units will be roughly \((1000 \times \text{Molar mass of glucose}) - (999 \times \text{Molar mass of water})\).
- Molar mass of polysaccharide \(\approx (1000 \times 180) - (999 \times 18) = 180000 - 17982 = 162018 \, \text{g/mol}\).
Osmolarity and Number of Particles
Osmolarity is proportional to the number of solute particles. A single molecule of the polysaccharide, even though it is large and made of many units, acts as one solute particle in solution. A single molecule of glucose also acts as one solute particle.
To have the same effect on osmolarity, we need to have the same number of moles of particles. We need to compare the number of moles in 1 gram of the polysaccharide to the number of moles in different masses of glucose.
Calculating Moles in 1 Gram Polysaccharide
The number of moles in 1 gram of the polysaccharide is:
\[ \text{Moles of polysaccharide} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1 \, \text{g}}{162018 \, \text{g/mol}} \]This is the number of osmolarity-contributing particles from the polysaccharide.
Calculating Moles in Glucose Options
Now let's calculate the number of moles for the different masses of glucose given in the options. Glucose has a molar mass of approximately \(180 \, \text{g/mol}\).
| Glucose Mass (mg) | Glucose Mass (g) | Moles of Glucose (\(\frac{\text{Mass}}{\text{Molar mass}}\)) |
|---|---|---|
| 1 mg | 0.001 g | \(\frac{0.001}{180}\) |
| 100 mg | 0.1 g | \(\frac{0.1}{180}\) |
| 500 mg | 0.5 g | \(\frac{0.5}{180}\) |
| 1000 mg | 1 g | \(\frac{1}{180}\) |
Comparing Osmolarity Effects
We are looking for the mass of glucose that provides approximately the same number of moles as 1 gram of the polysaccharide.
\[ \text{Moles of glucose} \approx \text{Moles of polysaccharide} \] \[ \frac{\text{Mass of glucose (g)}}{180} \approx \frac{1}{162018} \] \[ \text{Mass of glucose (g)} \approx \frac{180}{162018} \]Let's approximate the calculation:
\[ \text{Mass of glucose (g)} \approx \frac{180}{162000} = \frac{1}{900} \, \text{g} \]To convert grams to milligrams, we multiply by 1000:
\[ \text{Mass of glucose (mg)} \approx \frac{1}{900} \times 1000 = \frac{1000}{900} = \frac{10}{9} \approx 1.11 \, \text{mg} \]Comparing this calculated mass (approximately 1.11 mg glucose) to the given options:
- 1 mg glucose
- 100 mg glucose
- 500 mg glucose
- 1000 mg glucose
The value 1.11 mg is closest to 1 mg glucose.
Therefore, 1 gram of this polysaccharide has approximately the same osmolarity effect as 1 mg of glucose because they contain a similar number of moles of solute particles, even though their masses are very different.
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