An aqueous solution is composed of 7.50 g NaCl (MM = 58.44 g/mol) diluted to 0.100 L. Calculate the osmotic pressure of the solution at 298 K
a.
5.83 atm
b.
9.22 atm
c.
18.3 atm
d.
62.8 atm
To calculate the osmotic pressure of the solution, we can use the equation:
π = (n/V) * RT
where π is the osmotic pressure, n is the number of moles of solute, V is the volume of the solution in liters, R is the ideal gas constant (0.0821 L•atm/(mol•K)), and T is the temperature in Kelvin.
First, let's calculate the number of moles of NaCl in the solution:
n = mass / molar mass
n = 7.50 g / 58.44 g/mol
n = 0.1282 mol
Now, we can substitute the values into the osmotic pressure equation:
π = (0.1282 mol / 0.100 L) * (0.0821 L•atm/(mol•K)) * 298 K
π = 1.282 * 0.0821 * 298
π ≈ 31.5 atm
Therefore, the correct answer is not among the given options.
To calculate the osmotic pressure of the solution, we can use the following formula:
Osmotic pressure (π) = (n / V) * R * T
Where:
n = moles of solute
V = volume of solution in liters
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, let's calculate the moles of solute (NaCl).
Given:
Mass of NaCl = 7.50 g
Molar mass of NaCl (MM) = 58.44 g/mol
Moles of NaCl = mass / molar mass = 7.50 g / 58.44 g/mol
Next, we need to convert the volume of the solution to liters.
Volume of solution = 0.100 L
Now, let's substitute these values into the osmotic pressure formula:
π = (n / V) * R * T
π = (7.50 g / 58.44 g/mol) / 0.100 L * 0.0821 L·atm/(mol·K) * 298 K
By performing the calculations, we find that the osmotic pressure (π) is approximately 9.22 atm. Therefore, the correct answer is b. 9.22 atm.
pi = i*M*RT.
i is 2 for NaCl (the van't Hoff factor)
M = grams NaCl/molar mass/0.1L
R as usual
T given
Solve for pi in atm if R is 0.08206 L*atm/mol*K