Question

85% aqueous solution of nano3 is apparently 90% dissociated at 27 degree celcius calculate its osmotic pressure

Answer 1

Answer:

solution of NaNO3 means 0.85 gm of NaNO3 is dissolved in 100 g of solution or 100 cm3 = 100ml of the solution. Now, no. of moles of NaNO3 in given sample = given mass/ molar mass = 0.85 / 85 (molar mass of NaNO3) = 0.01 moles. Putting the values we get the osmotic pressure as: P = 0.01/0.1 x 0.0821 x 300 = 2.463 atm