A mixture of NaCl and sucrose (C12H22O11) of combined mass 10.2 g is dissolved in enough water to make up a 250 mL solution. The osmotic pressure of the solution is 8.29 atm at 23°C. Calculate the mass percent of NaCl in the mixture.
To calculate the mass percent of NaCl in the mixture, we need to determine the moles of NaCl and sucrose in the solution first.
1. Calculate the moles of sucrose:
Mass of C12H22O11 = 10.2 g
Molar mass of C12H22O11 = 342.30 g/mol
Moles of C12H22O11 = (mass of C12H22O11) / (molar mass of C12H22O11)
Moles of C12H22O11 = 10.2 g / 342.30 g/mol
2. Calculate the moles of NaCl using the osmotic pressure:
Osmotic pressure = π = (n/V)RT
where:
n = moles of solute
V = volume of solution (in liters) = 250 mL = 0.250 L
R = gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin = 23°C + 273.15 = 296.15 K
Rearranging the equation, we can solve for moles of solute:
n = (πV) / (RT)
Moles of NaCl = (8.29 atm)(0.250 L) / (0.0821 L·atm/(mol·K))(296.15 K)
3. Calculate the mass percent of NaCl:
Mass of NaCl = (moles of NaCl) x (molar mass of NaCl)
Mass percent = (mass of NaCl / mass of mixture) x 100
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl = (moles of NaCl) x (molar mass of NaCl)
Mass percent of NaCl = (mass of NaCl / mass of mixture) x 100
Now, let's plug in the values to calculate:
Mass of C12H22O11:
= 10.2 g
Molar mass of C12H22O11:
= 342.30 g/mol
Moles of C12H22O11:
= 10.2 g / 342.30 g/mol
Osmotic pressure (π):
= 8.29 atm
Volume of solution (V):
= 250 mL = 0.250 L
Gas constant (R):
= 0.0821 L·atm/(mol·K)
Temperature (T):
= 23°C + 273.15 = 296.15 K
Moles of NaCl:
= (8.29 atm)(0.250 L) / (0.0821 L·atm/(mol·K))(296.15 K)
Molar mass of NaCl:
= 58.44 g/mol
Mass of NaCl:
= (moles of NaCl) x (molar mass of NaCl)
Mass percent of NaCl:
= (mass of NaCl / mass of mixture) x 100
This will give us the final answer: the mass percent of NaCl in the mixture.
To calculate the mass percent of NaCl in the mixture, we need to determine the amount of NaCl and sucrose present in the solution.
First, we need to find the moles of NaCl and sucrose using the ideal gas law:
PV = nRT
Where:
P = osmotic pressure (8.29 atm)
V = volume of the solution (250 mL)
n = moles of solute
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (23 + 273 = 296 K)
Rearranging the ideal gas equation to solve for moles:
n = PV / RT
For NaCl:
n(NaCl) = (8.29 atm) × (0.250 L) / (0.0821 L·atm/mol·K) × (296 K)
Next, we calculate the moles of sucrose using the same formula:
n(C12H22O11) = (8.29 atm) × (0.250 L) / (0.0821 L·atm/mol·K) × (296 K)
To convert the moles of NaCl and sucrose to grams, we use their respective molar masses:
Molar mass of NaCl = 58.44 g/mol
Molar mass of C12H22O11 = (12.01 × 12) + (1.01 × 22) + (16.00 × 11) = 342.3 g/mol
Now, we can calculate the mass percent of NaCl:
mass percent of NaCl = (mass of NaCl / total mass of solution) × 100
The mass of NaCl can be found using the moles of NaCl and its molar mass:
mass of NaCl = moles of NaCl × molar mass of NaCl
Finally, the total mass of the solution is the sum of the masses of NaCl and sucrose:
total mass of solution = mass of NaCl + mass of sucrose
By substituting the values into the formula, we can find the mass percent of NaCl in the mixture.
This a multistep problem involving simultaneous equations.
pi = i*M*R*T. I is the van't Hoff factor. Substitute and solve for M = molarity.
Then you have two equations.
Let X = grams NaCl
and Y = grams sucrose = SU
equation 1 is X +Y = 10.2
Then mols NaCl + mols SU = total mols so equation 2 is
(X/58.5) + (Y/342) = 0.250 x M (from pi calculation)
Solve the two equations for X and Y.
Then %X = (mass X/10.2)*100 = ?