Question

Determine the osmotic pressure of a solution that contains 0.048 g of a hydrocarbon solute (molar mass = 340 g/mol0

Answer 1

The osmotic pressure of a solution containing 0.048 g of a hydrocarbon with a molar mass of 340 g/mol is calculated by first finding the molarity and then using the osmotic pressure formula, resulting in approximately 0.00345 atm.

Step-by-step explanation:

To determine the osmotic pressure of a solution containing a hydrocarbon solute, first, calculate the molarity of the solution. The molarity (M) is the number of moles of solute per liter of solution. Since the molar mass of the hydrocarbon is given as 340 g/mol, you can find the number of moles of solute by dividing the mass of the solute by its molar mass.

Number of moles = Mass (g) / Molar Mass (g/mol) = 0.048 g / 340 g/mol = 0.000141 mol

The volume of the solution is not given, but assuming it's 1 liter for the sake of calculation. Divide the number of moles by the volume to get molarity:

Molarity (M) = Number of moles / Volume (L) = 0.000141 mol / 1 L = 0.000141 M

To find the osmotic pressure, use the formula:

Osmotic Pressure (II) = MRT

Where M is the molarity, R is the gas constant (0.0821 L·atm/(K·mol)), and T is the absolute temperature in Kelvin. Assuming room temperature (25°C or 298 K) for the calculation:

II = (0.000141 M) x (0.0821 L·atm/K·mol) x (298 K)

Therefore, the osmotic pressure is:

II = (0.000141) x (0.0821) x (298) ≈ 0.00345 atm

Answer 2

m(CH) = 0.048 g
M(CH) = 340 g/mol
V = 350 mL = 0.35 L
T = 20 C = 293 K
π = ?

For determination of osmotic pressure we will use next equation:

π = c x R x T

c = n/V

n= m/M = 0.048/340= 0.00014 mol

c = 0.00014/0.35 = 0.0004 mol/L

π = 0.0004 mol/L × 8.314 LkPa/Kmol × 293 K

π = 0.97 kPa