The volume of an aqueous solution of NaCl at 25oC is expressed as a function of the
amount m in 1kg of solvent in the following form;
V/cm3 = 1000.94 + 16.4 (m/mol) + 2.14 (m/mol)3/2
- 0.0027 (m/mol)5/2
Find the partial molar volume of NaCl and water in 1 mol kg-1
solution.
Why did the NaCl solution go to therapy? Because it was feeling a little salty!
To find the partial molar volume of NaCl and water in a 1 mol kg-1 solution, we can take the derivative of the volume equation with respect to the amount of NaCl (m).
First, let's find the partial molar volume of NaCl:
Partial molar volume of NaCl = ∂V/∂m
∂V/∂m = 16.4 + (2.14 * 1.5 * √m) - (0.0027 * 2.5 * √m)
Next, let's find the partial molar volume of water:
Partial molar volume of water = ∂V/∂m * (1 - m)
Now, let's put on our clown disguises and solve the equations!
To find the partial molar volume of NaCl and water in a 1 mol kg-1 solution, we need to take the derivative of the volume expression with respect to the amount of the substance.
First, let's find the partial molar volume of NaCl:
1. Take the derivative of the volume expression with respect to the amount of NaCl, m:
dV/dm = 16.4 + (2.14 * 3/2 * sqrt(m)) - (0.0027 * 5/2 * sqrt(m))
2. Substitute m = 1 mol and evaluate the derivative:
dV/dm = 16.4 + (2.14 * 3/2 * sqrt(1)) - (0.0027 * 5/2 * sqrt(1))
dV/dm = 16.4 + (2.14 * 3/2 * 1) - (0.0027 * 5/2 * 1)
dV/dm = 16.4 + 3.21 - 0.00675
dV/dm = 19.60325
So, the partial molar volume of NaCl in a 1 mol kg-1 solution is approximately 19.60325 cm3/mol.
Next, let's find the partial molar volume of water:
1. Take the derivative of the volume expression with respect to the amount of water, m' (assuming 1 mol of NaCl in 1 kg of solvent, so m' relates to the amount of water):
dV/dm' = 0
Since the derivative with respect to m' is 0, the partial molar volume of water is not influenced by the amount of water.
Therefore, the partial molar volume of water in a 1 mol kg-1 solution is 0 cm3/mol.
To find the partial molar volume of NaCl and water in a 1 mol kg-1 solution, we will first differentiate the given volume expression with respect to the amount of NaCl (m):
V = 1000.94 + 16.4m + 2.14m^(3/2) - 0.0027m^(5/2)
The partial molar volume of NaCl, denoted as V̄_NaCl, is defined as the derivative of the volume with respect to the amount of NaCl at constant temperature and pressure. Mathematically, it can be expressed as:
V̄_NaCl = (∂V/∂m)_T,P
To find V̄_NaCl, we need to take the derivative of the volume expression with respect to m:
∂V/∂m = 16.4 + (3/2) * 2.14 * √m - (5/2) * 0.0027 * m^(3/2)
Now, we can substitute m = 1 mol kg-1 into the expression for ∂V/∂m to find the partial molar volume of NaCl:
V̄_NaCl = (∂V/∂m) = 16.4 + (3/2) * 2.14 * √1 - (5/2) * 0.0027 * 1^(3/2)
Simplifying this expression will give us the partial molar volume of NaCl in cm^3 mol-1.
Similarly, we can find the partial molar volume of water, denoted as V̄_water, by differentiating the volume expression with respect to the amount of water (1 - m):
V̄_water = (∂V/∂(1-m))_T,P
To find V̄_water, we need to substitute (1 - m) into the volume expression and differentiate it with respect to (1 - m), then substitute m = 1 mol kg-1 into the resulting expression.
Note: The partial molar volume of water can also be determined by subtracting the partial molar volume of NaCl (V̄_NaCl) from the total molar volume of the solution at the given concentration.
By following these steps, we can calculate the partial molar volumes of NaCl and water in the 1 mol kg-1 solution.