Answer to Question #141495 in Chemistry for ivy

Question

The non-volatile, non-electrolyte TNT (trinitrotoluene), C7H5N3O6
(227.10g/mol), is soluble in benzene C6H6. Calculate the osmotic
pressure generated when 14.3g of TNT are dissolved in 242ml of a
benzene solution at 298 K.

Answer 1

Solution:

The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.

Π = i × C_M × R × T

where:

• Π is the osmotic pressure,
• i is the van't Hoff factor, 
• C_M is the molar concentration,
• R is the ideal gas constant (0.0821 L atm mol^-1 K^-1),
• T is the temperature in Kelvin.

Molar concentration (C_M) is the number of moles of solute per liter of solution, which can be calculated using the following equation: C_M = mol solute / L of solution.

C_M = n / V = m / (M × V)

C_M(TNT) = (14.3 g) / (227.10 g mol^-1 × 0.242 L) = 0.2602 mol L^-1 = 0.26 mol L^-1

C_M(TNT) = 0.26 mol L^-1

The van't Hoff factor (i) for the solute is 1 because it is a nonelectrolyte.

Hence,

Π = C_M × R × T

Π = 0.26 mol L^-1 × 0.0821 L atm mol^-1 K^-1 × 298 K = 6.36 atm

Π = 6.36 atm

Answer: the osmotic pressure (Π) is 6.36 atm.