(Due date: February ,2018) 1) Calculate the molality of each of the following aq

Question

(Due date: February ,2018) 1) Calculate the molality of each of the following aqueous solutions: (a) 0.940 M KBr solution (density of solution 1.08 g/mL). (b) 29.2 percent by mass NaCl solution, (c) 14.3 g of sucrose (C12H201) in 676 g of water 2) The solubility of N2 in blood at 37°C and at a partial pressure of 0.80 atm is 5.6 × 10-4 mol L. A diver breathes compressed air with the partial pressure of N equal to 4.5 atm. Calculate the N2 are in the blood of the diver blood in the body is 5.0L deep-sea number of moles of when breathing at this level in the deep-sea. Assume that the total volume of 3) How many grams of urea [(NH)-CO] must be added to 500 g of water to give a solution with a vapor Hg less than that of pure water at 30°C? (The vapor pressure of water at 30°C is 31.8 mmg.) 4) The solubility of KNO, i155g er 100 g of water at 75 C and 38.0 g at 25 C. What mass in grams) of KN Os will crystallize out of solution if exactly 150 g of its saturated solution at 75°C is cooled to 25°C? 5) In the experiment of using freezing-point depression to find the molar mass of a compound, a solution of 3.80 g of a compound having the empirical formula CaHlsP in 19.6 g of benzene is observed to free Calculate the molar mass of the solute and its molecular formula. (For benzene, the normal f 5.5 C and Kr value is 5.12°C/m.) eezing point is 6) A solution containing 0.8330 g of a polymer of unknown structure in 170.0 mL of an organic solve found to have an osmotic pressure of 5.20 mmllg at 25°C. Determine the molar mass of the polymer. (Hint: For polymers, the van't Hoff factor () is equal to I. nt was C.Determine the molar mass of the polymer. (Hint: For

Explanation / Answer

Ans 1 :

a) 0.940 M KBr Solution

Let the volume of solution be 1 L

So number of moles of KBr = 0.940 mol

mass of solute = 0.940 x molar mass KBr

= 0.940 x 119.002 g/mol

= 111.86 g

Density = mass / volume

1.08 = mass / 1

Mass = 1.08 kg or 1080 g

Mass of solvent = 1080 g - 111.86 g = 968.14 g or 0.96814 kg

molality = no. of moles of solute / mass of solvent in kg

= 0.940 / 0.96814

= 0.97 m

b) 29.2 percent NaCl solution

Let the mass of solution be 100 g

So mass of NaCl = 29.2 g

Mass of solvent = 100 - 29.2 = 70.8 g

Number of mol NaCl = 29.2 / 58.4 = 0.5 mol

Molality = mol / mass of solvent in kg

= 0.5 / 0.0708

= 7.06 m

c) 14.3 g of sucrose in 676 g water

Number of moles of sucrose = 14.3 / 342.3

= 0.0418 mol

Molality = no. of mole / mass of solvent in kg

= 0.0418 / 0.676

= 0.062 m

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