To calculate the osmolarity of the solution, we need to consider the contributions of both dextrose and sodium chloride.

Step 1: Calculate the Molarity of Dextrose

• Dextrose concentration = 7.5% (w/v), meaning 7.5 grams of dextrose in 100 mL of solution.
• Molecular weight (M.W.) of dextrose (C₆H₁₄O₇) = 180 g/mol.

Molarity (M) of dextrose:

$\text{Molarity of dextrose} = \frac{7.5 \text{ g}}{180 \text{ g/mol} \times 0.1 \text{ L}} = \frac{7.5}{18} \text{ mol/L} = 0.4167 \text{ mol/L}$

Step 2: Calculate the Molarity of Sodium Chloride (NaCl)

• Sodium chloride concentration = 0.9% (w/v), meaning 0.9 grams of NaCl in 100 mL of solution.
• Molecular weight (M.W.) of NaCl = 58.4 g/mol.

Molarity (M) of sodium chloride:

$\text{Molarity of NaCl} = \frac{0.9 \text{ g}}{58.4 \text{ g/mol} \times 0.1 \text{ L}} = \frac{0.9}{5.84} \text{ mol/L} = 0.1541 \text{ mol/L}$

Step 3: Determine the Number of Particles in Solution

• Dextrose is a non-electrolyte, so it does not dissociate in solution. Therefore, it contributes 1 particle per molecule.
• NaCl dissociates into 2 particles (Na⁺ and Cl⁻) in solution.

Step 4: Calculate the Osmolarity

• Osmolarity of dextrose = 0.4167 mol/L × 1 = 0.4167 osmoles/L.
• Osmolarity of NaCl = 0.1541 mol/L × 2 = 0.3082 osmoles/L.

Step 5: Add the Osmolarities

$\text{Total osmolarity} = 0.4167 \text{ osmoles/L} + 0.3082 \text{ osmoles/L} = 0.7249 \text{ osmoles/L}$

Thus, the osmolarity of the solution is approximately 0.725 osmoles/L.

Would you like further details or have any other questions?

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Here are some related questions:

1. How does osmolarity differ from osmolality?
2. What are the effects of osmolarity on cellular function?
3. How do you calculate the osmolarity of a solution with multiple electrolytes?
4. Why does sodium chloride dissociate into two particles in solution?
5. What is the significance of molecular weight in calculating osmolarity?

Tip: Always double-check the dissociation of compounds when calculating osmolarity, as different solutes may dissociate differently in solution.