A 250 -mL bottle of a sports drink solution contains by mass of sodium chloride. What is the molal concentration of sodium chloride in this bottle of sports drink?
Molar mass
58.44
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(B)
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The Correct Answer and Explanation is:
To solve this problem, we need to determine the molal concentration (molality) of sodium chloride (NaCl) in the sports drink. Molality is defined as the number of moles of solute per kilogram of solvent. The formula for molality (m) is: m=moles of solutemass of solvent (kg)m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}
Given Information:
- Volume of the sports drink solution: 250 mL
- Mass percentage of sodium chloride: This seems to be missing in the problem. For the sake of the explanation, I’ll assume it to be 5% by mass (you can adjust the steps accordingly if the correct percentage is provided).
- Molar mass of sodium chloride (NaCl): 58.44 g/mol
- Density of the solution: Since this is not provided, we will assume it is close to the density of water (1 g/mL), so 250 mL of solution would weigh approximately 250 grams.
Steps to Calculate Molality:
- Calculate the mass of sodium chloride (NaCl):
If the sports drink is 5% NaCl by mass, we calculate the mass of NaCl in 250 grams of solution. Mass of NaCl=250 g×5100=12.5 g\text{Mass of NaCl} = 250 \, \text{g} \times \frac{5}{100} = 12.5 \, \text{g} - Convert the mass of NaCl into moles:
Using the molar mass of sodium chloride (58.44 g/mol), we calculate the number of moles of NaCl: Moles of NaCl=12.5 g58.44 g/mol≈0.214 mol\text{Moles of NaCl} = \frac{12.5 \, \text{g}}{58.44 \, \text{g/mol}} \approx 0.214 \, \text{mol} - Determine the mass of the solvent (water):
Since the total mass of the solution is 250 g and the mass of NaCl is 12.5 g, the mass of the solvent (water) is: Mass of water=250 g−12.5 g=237.5 g=0.2375 kg\text{Mass of water} = 250 \, \text{g} – 12.5 \, \text{g} = 237.5 \, \text{g} = 0.2375 \, \text{kg} - Calculate the molality:
Finally, we can calculate the molality of NaCl using the formula for molality: m=0.214 mol0.2375 kg≈0.902 mol/kgm = \frac{0.214 \, \text{mol}}{0.2375 \, \text{kg}} \approx 0.902 \, \text{mol/kg}
Conclusion:
The molal concentration of sodium chloride in the sports drink solution is approximately 0.902 mol/kg.
Please check if the mass percentage of NaCl or the volume of the solution is different, as the molality depends on these values. If more precise details are provided (e.g., exact percentage of NaCl), you can adjust the calculation accordingly.