What is the boiling-point change for a solution containing 0.347 mol of naphthal

Question

What is the boiling-point change for a solution containing 0.347 mol of naphthalene (a nonvolatile, nonionizing compound) in 250. g of liquid benzene? (K_b = 2.53 degree C/m benzene) a) 3.51 degree C b) 1.82 degree C c) 7.29 degree C d) 0.878 degree C e) 0.219 degree C When a 41.7-g sample of an unknown compound is dissolved in 500 g of benzene, the freezing point of the resulting solution is 3.77 degree C. The freezing point of pure benzene is 5.48 degree C, and A for benzene is 5, 12 degree C/m. Calculate the molar mass of the unknown compound. a) 124.9 g/mol b) 24.4 g/mol c) 214 g/mol d) 499 g/mol e) 250. g/mol A 2.2-g sample of a small protein having a molecular weight of 42,000 g/mol is dissolved in 45.6 mL of water at 23 degree C. What is the osmotic pressure of the solution? (R = 0.0821 L middot atm/(K middot mol)) a) 0.028 mmHg b) 1.6 mmHg c) 21 mmHg d) 27000 mmHg e) 890 mmHg What is the freezing point of an aqueous 1.18 m CaCl_2, solution? (k_f for water is 1.858 degree C/m.) a) -2.2 degree C b) 2.2 degree C c) -6.6 degree C d) -6.6 degree C e) 0.0 degree C.

Explanation / Answer

Q11.

Apply colligative properties

dTb = Kb*m

Kb = 2.53 C/molal

m = molality

m = mol of solute / kg solvent

mol of solute = 0.347

kg solvent = 250 g = 0.250 kg

so

m = 0.347/0.25 = 1.388

so

dTb = 1.388*2.53

dTb = 3.51164

best answer is A)

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