Let's solve the problem step by step: ### **Given:** - **Concentration:** 8.4% w/v (which means 8.4 g per 100 mL, or 84 g per 1 L) - **Molecular Weight (MW) of NaHCO₃:** 84 g/mol ### **Step 1: Find Molarity** - 84 g NaHCO₃ in 1 L solution - Moles of NaHCO₃ = \( \frac{84\text{ g}}{84\text{ g/mol}} = 1\text{ mol} \) - Molarity = 1 mol/L ### **Step 2: Find Osmolarity** - NaHCO₃ dissociates into 2 particles in solution: Na⁺ and HCO₃⁻ - So, 1 mol of NaHCO₃ gives 2 osmoles Osmolarity = 1 mol/L × 2 = 2 osmol/L = 2000 mOsmol/L --- ### **Final Answer** **B) 2,000 mOsmol/L**

Let's carefully work through the calculation of the osmolarity of an 8.4% w/v sodium bicarbonate (NaHCO₃) solution step by step. --- ### **Step 1: Understand the given data** - **Concentration:** 8.4% w/v - This means **8.4 grams of NaHCO₃ per 100 mL** of solution. - **Molecular weight (MW) of NaHCO₃:** 84 g/mol --- ### **Step 2: Convert percentage to molarity** - To find molarity, first determine how many grams of NaHCO₃ are present in 1 liter (1000 mL): \[ \text{grams per liter} = \frac{8.4 \text{ g}}{100 \text{ mL}} \times 1000 \text{ mL} = 84 \text{ g} \] - **Number of moles of NaHCO₃ in 1 liter:** \[ \text{Moles} = \frac{\text{grams}}{\text{MW}} = \frac{84 \text{ g}}{84 \text{ g/mol}} = 1 \text{ mol} \] - **Thus, the molarity (M) of the solution:** \[ \boxed{ \text{Molarity} = 1 \text{ mol/L} } \] --- ### **Step 3: Determine the number of particles (osmoles) produced upon dissociation** - Sodium bicarbonate dissociates in water as: \[ \text{NaHCO}_3 \rightarrow \text{Na}^+ + \text{HCO}_3^- \] - **Number of particles per molecule:** - 1 molecule yields 2 ions - **Osmolarity:** \[ \text{Osmolarity} = \text{Molarity} \times \text{number of particles per molecule} = 1 \text{ mol/L} \times 2 = 2 \text{ osmoles/L} \] - **Expressed in milliosmoles per liter (mOsmol/L):** \[ 2 \text{ osmol/L} = 2000 \text{ mOsmol/L} \] --- ### **Final Answer:** **The osmolarity of the 8.4% w/v sodium bicarbonate solution is approximately:** \[ \boxed{ 2000 \text{ mOsmol/L} } \] This corresponds to option B. --- ### **Summary:** - 8.4% w/v NaHCO₃ solution contains 84 g per liter. - Moles of NaHCO₃ per liter = 1 mol. - Dissociation produces 2 particles per molecule. - Therefore, osmolarity = 2 osmol/L = 2000 mOsmol/L. **Hence, the osmolarity is approximately 2000 mOsmol/L.**