Why is the molality of solution made by adding 9.5g of NaCl to 300g of water
The Correct Answer and Explanation is:
Question (interpreted):
What is the molality of the solution made by adding 9.5 g of NaCl to 300 g of water? Provide the correct answer and a detailed explanation (around 300 words).
Step 1: Understanding Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula is: molality(m)=moles of solutekg of solvent\text{molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}}
- Solute: NaCl (sodium chloride)
- Solvent: Water (H₂O)
- Given:
- Mass of NaCl = 9.5 g
- Mass of water = 300 g = 0.300 kg
Step 2: Calculate moles of NaCl
The molar mass of NaCl is: Molar mass of NaCl=(22.99 g/mol for Na)+(35.45 g/mol for Cl)=58.44 g/mol\text{Molar mass of NaCl} = (22.99 \text{ g/mol for Na}) + (35.45 \text{ g/mol for Cl}) = 58.44 \text{ g/mol}
Calculate moles of NaCl: moles of NaCl=9.5 g58.44 g/mol≈0.1625 mol\text{moles of NaCl} = \frac{9.5 \text{ g}}{58.44 \text{ g/mol}} \approx 0.1625 \text{ mol}
Step 3: Calculate molality
The mass of solvent (water) in kilograms: 300 g=0.300 kg300 \text{ g} = 0.300 \text{ kg}
Now calculate molality: m=0.1625 mol0.300 kg=0.5417 mol/kgm = \frac{0.1625 \text{ mol}}{0.300 \text{ kg}} = 0.5417 \text{ mol/kg}
Final answer:
Molality = 0.54 mol/kg (rounded to two decimal places)
Explanation
Molality is a concentration unit used to express the amount of solute present in a given amount of solvent. Unlike molarity, which depends on the volume of solution, molality depends only on the mass of the solvent, making it temperature independent. This makes molality especially useful in thermodynamic calculations.
In this problem, we are asked to find the molality of a solution prepared by dissolving 9.5 grams of NaCl in 300 grams of water. To determine the molality, we first need to find the number of moles of NaCl. This requires the molar mass of NaCl, which is calculated by adding the atomic masses of sodium (Na, approximately 23 g/mol) and chlorine (Cl, approximately 35.45 g/mol), yielding 58.44 g/mol.
By dividing the mass of NaCl (9.5 g) by its molar mass (58.44 g/mol), we get approximately 0.1625 moles of NaCl. Next, molality is calculated by dividing the moles of solute by the mass of solvent in kilograms. Since 300 grams of water is equivalent to 0.300 kilograms, dividing 0.1625 moles by 0.300 kg gives a molality of about 0.54 mol/kg.
This value indicates that there are 0.54 moles of NaCl dissolved per kilogram of water in the solution. Molality is a useful measurement in many practical applications, including colligative property calculations (like boiling point elevation and freezing point depression), because it is unaffected by temperature changes that can alter solution volume.
Thus, the molality of the solution made by adding 9.5 grams of NaCl to 300 grams of water is 0.54 mol/kg.
