The solubility of sodium chloride is 36.0 g/100. ml of water, and the density of saturated NaCl solution is 1.202 g/mL.. I have 500. mL of a 1.00 M NaCl solution. If I boil it to the point of saturation, what is the final volume?
To determine the final volume of a 1.00 M NaCl solution when it is boiled to the point of saturation, we need to calculate the mass of NaCl dissolved in the solution.
Given:
- Solubility of NaCl = 36.0 g/100 mL of water
- Density of saturated NaCl solution = 1.202 g/mL
- Initial volume = 500 mL
- Initial molarity = 1.00 M
First, we can calculate the initial mass of NaCl in the 500 mL solution:
mass = volume × density
mass = 500 mL × 1.202 g/mL = 601 g
Next, we can determine the number of moles of NaCl in the initial solution using the molarity and volume:
moles = molarity × volume
moles = 1.00 M × 0.500 L = 0.500 moles
Since the solubility is given in terms of mass per volume, we can calculate how much water is needed to dissolve this amount of NaCl:
mass NaCl = (36.0 g/100 mL) × V final
0.500 moles NaCl × (58.44 g/mol) = (36.0 g/100 mL) × V final
29.22 g = (36.0 g/100 mL) × V final
V final = 100 × (29.22 g / 36.0 g)
V final = 81.17 mL
Thus, when the 1.00 M NaCl solution is boiled to the point of saturation, the final volume will be approximately 81.17 mL.
To determine the final volume of the NaCl solution after boiling it to the point of saturation, you need to consider the solubility of sodium chloride and the density of the saturated solution.
First, let's calculate the amount of sodium chloride needed to make a 1.00 M NaCl solution with a volume of 500 mL.
Molarity (M) is defined as moles of solute per liter of solution. You have a 1.00 M NaCl solution with a volume of 500 mL (0.500 L). Therefore, the moles of sodium chloride (NaCl) can be calculated as follows:
Moles = Molarity x Volume
Moles = 1.00 mol/L x 0.500 L
Moles = 0.500 moles
Since the molar mass of NaCl is 58.44 g/mol, we can calculate the mass of NaCl in the solution using:
Mass = Moles x Molar mass
Mass = 0.500 moles x 58.44 g/mol
Mass = 29.22 grams
Now, let's move on to boiling the solution to the point of saturation. The solubility of NaCl in water is given as 36.0 g/100 mL of water. This means that at the boiling point, 100 mL of water can dissolve 36.0 grams of NaCl.
To find the final volume, you need to determine how much water is needed to dissolve the initial mass of NaCl (29.22 grams) when boiled to the point of saturation.
Volume of water needed = (Mass of NaCl / Solubility of NaCl in water) x 100 mL
Volume of water needed = (29.22 g / 36.0 g/100 mL) x 100 mL
Volume of water needed = 81.17 mL
Finally, to calculate the final volume of the solution after boiling, you add the volume of water needed to the initial volume of the NaCl solution:
Final volume = Initial volume + Volume of water needed
Final volume = 500 mL + 81.17 mL
Final volume = 581.17 mL
Therefore, the final volume of the NaCl solution after boiling it to the point of saturation is approximately 581.17 mL.