What are the boiling point and freezing point of a_5.47 = solution of naphthalen

Question

What are the boiling point and freezing point of a_5.47 = solution of naphthalene in benzene? (The boiling point and freezing point of benzene are 80.1 degree C and 5.5 degree respectively. The boiling point elevation constant for _benzene is 2.53 degree C/m, and the freezing point depression constant for _benzene is 5.12 degree C/m.) Boiling point = _____ degree C Freezing point = ______ degree C Enter your answer in the provided box. A 4.10-g sample of a salt dissolves in 8.80 g of water to give a saturated solution at 22 degree C. What is the solubility (in g salt/100 g of H_2O) of the salt? ___ g salt/100 g Enter your answer in the provided box. The osmotic pressure of 7.64 times 10^-3 M solutions of CaCl_2 and urea at 25 degree C are 0.462 and_0.187 atm, respectively. Calculate the van't Hoff factor for the CaCl_2 solution. i = ___ Enter your answer in the provided box. What is the osmotic pressure (in atm) of a 0.985 M urea solution at 11 degree C? ____ atm

Explanation / Answer

1) T = mKb = 5.47*2.53 = 13.8391 oC
   Boiling point of the solution = 80.1 + 13.8391 = 93.9391oC
     T = mKf = 5.47 * 5.12 = 28.0064 oC
   Freezing point of the solution = 5.5 - 28.0064 = -22.5064oC

2) The ratio of two osmotic pressures will give i for CaCl2 solution,
   i = 0.462 / 0.187 = 2.4705

3) 4.10 gm salt is soluble in 8.80gm of water
   x gm of salt will be soluble in 100 gm of water
Just cross-multiply to get the answer, (4.10 * 100) / 8.80 = 46.5909 gm salt / 100 g of H2O

4) osmotic pressure, = MRT = 0.985 mol/L * 0.08206 L atm mol-1K-1 * 284.15 K = 22.9675 atm

   

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