Give the mass of the solute and mass of the solvent for 219 g of a solution that is 0.55 m in Na2CO3 , starting with the solid solute.
The correct answer and explanation is:
Let’s solve this step-by-step:
Problem Recap:
- Solution mass = 219 g
- Molality (m) of Na₂CO₃ = 0.55 m (mol/kg solvent)
- Need to find:
- Mass of solute (Na₂CO₃)
- Mass of solvent (water)
Step 1: Understand molality
Molality (m) = moles of solute / kg of solvent
Given: m = 0.55 mol/kg solvent
Step 2: Define variables
Let:
- msolventm_{solvent} = mass of solvent (in kg)
- msolutem_{solute} = mass of Na₂CO₃ (in g)
- Total mass =msolute+msolvent×1000=219= m_{solute} + m_{solvent} \times 1000 = 219 g
Step 3: Use molality formula to relate moles solute and solvent
Moles of solute = 0.55 × msolventm_{solvent} (in kg)
Step 4: Calculate molar mass of Na₂CO₃
- Na = 22.99 g/mol
- C = 12.01 g/mol
- O = 16.00 g/mol × 3 = 48.00 g/mol
Sum: 2×22.99+12.01+48.00=45.98+12.01+48.00=105.992 \times 22.99 + 12.01 + 48.00 = 45.98 + 12.01 + 48.00 = 105.99 g/mol
Step 5: Express mass of solute in terms of solvent mass
Mass of solute = moles × molar mass = 0.55×msolvent×105.990.55 \times m_{solvent} \times 105.99
Since msolventm_{solvent} is in kg, mass of solute in grams = 0.55×msolvent×105.990.55 \times m_{solvent} \times 105.99
Step 6: Write total mass equation
Total mass = mass solute + mass solvent
219 = 0.55×msolvent×105.99+msolvent×10000.55 \times m_{solvent} \times 105.99 + m_{solvent} \times 1000
Step 7: Solve for msolventm_{solvent} (in kg)
219=0.55×105.99×msolvent+1000×msolvent219 = 0.55 \times 105.99 \times m_{solvent} + 1000 \times m_{solvent}
Calculate coefficient:
0.55×105.99=58.29450.55 \times 105.99 = 58.2945
So:
219=58.2945×msolvent+1000×msolvent219 = 58.2945 \times m_{solvent} + 1000 \times m_{solvent}
219=(58.2945+1000)×msolvent219 = (58.2945 + 1000) \times m_{solvent}
219=1058.2945×msolvent219 = 1058.2945 \times m_{solvent}
msolvent=2191058.2945≈0.2069 kgm_{solvent} = \frac{219}{1058.2945} \approx 0.2069 \, \text{kg}
Step 8: Calculate mass of solute
Mass solute = 0.55×0.2069×105.990.55 \times 0.2069 \times 105.99
=0.55×0.2069×105.99≈12.06 g= 0.55 \times 0.2069 \times 105.99 \approx 12.06 \, \text{g}
Step 9: Calculate mass of solvent in grams
msolvent=0.2069×1000=206.9 gm_{solvent} = 0.2069 \times 1000 = 206.9 \, \text{g}
Final answer:
- Mass of Na₂CO₃ (solute) = 12.06 g
- Mass of solvent (water) = 206.9 g
Explanation (approx. 300 words):
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this problem, a 0.55 molal solution of Na₂CO₃ means there are 0.55 moles of sodium carbonate dissolved in every 1 kilogram of water. The total mass of the solution is given as 219 grams, which is the sum of the solute and solvent masses.
To find the individual masses, we first define variables for the solvent and solute mass. The molality allows us to express moles of solute in terms of the solvent mass. Next, by calculating the molar mass of Na₂CO₃ (approximately 105.99 g/mol), we can convert moles of solute to grams. This step is essential because the problem asks for the mass of the solute in grams.
Setting up the total mass equation: the sum of solute and solvent masses equals the total solution mass (219 g). This equation involves one unknown, the solvent mass, which we solve for. After obtaining the solvent mass in kilograms, we multiply by 1000 to convert to grams for easier interpretation.
Finally, substituting back, we find the solute mass in grams. The calculations reveal that out of the 219 g solution, approximately 12.06 g is sodium carbonate and about 206.9 g is water.
This method highlights how molality relates to mass and moles, a critical concept in chemistry for preparing solutions with precise concentrations. The approach ensures correct stoichiometric ratios and is especially useful when temperature changes affect volume but not mass.