Question

The nonvolatile, nonelectrolyte TNT (trinitrotoluene), C7H5N3O6 (227.1 g/mol), is soluble in benzene C6H6. How many grams of TNT are needed to generate an osmotic pressure of 5.14 atm when dissolved in 279 ml of a benzene solution at 298 K. grams TNT

Answer 1

Answer:

Explanation:

formula of osmotic pressure is as follows

p= n RT

n is mole of solute per unit volume

If m be the grams of solute needed

m gram = m / 227.1 moles

m / 227.1 moles dissolved in .279 litres

n = m / (227.1 x .279 )

= m / 63.36

substituting the values in the osmotic pressure formula

5.14 = (m / 63.36)  x .082 x 298

m / 63.36 = .21

m = 13.32 grams .