1. How many grams of calcium Chloride would be requird to be added to 500.0 gram

Question

1. How many grams of calcium Chloride would be requird to be added to 500.0 grams of water to ensure the freezing point would be -2.5 degrees C? Given Kfim = T

2. iff 222.8 grams of a non-volitile molecular compound lowers the vapor pressure of pure water by 3.0 mmHg when made into 375.0 grams of water. What would be the Molar Mass of the unknown compound given the vapor pressure of pure water at 30.0 degree C is 31.8 mmHg. (P(vapor pressure of solution)=X(mole fraction of solvent)Po(vapore pressure of pure solvent))

3. Dr. Himes cut his finger on a nail and got sepsis meaning an infection began at the sight of the cut and traveled up his arm going to his heart. He had to go to the ER where they gave this 75 Kg professor 25 mm per Kg of his body mass of Vancomycin (C66H75Cl2N9O24 MM = 1,449.3 g/mol). Please give the concentration of total Vancomycin in a 2.0 L IV bag in the terms below given the density is 10 g/mL.

Molarity________

Molality________

Mass Percent________

Mole Fraction_________

Explanation / Answer

Post one more question to get the answer to the third problem

Q1) Depression in Freezing Point = i * Kf * m

2.5 = 3(vont hoff factor for CaCl2) * 1.86 * m

m = 0.4480 m

0.4480 = number of moles of solute/volume of solvent in Kg

number of moles of solute = 0.2240 moles

Molar mass of CaCl2 = 40 + 2 * 35.5 = 111 gm/mol

Mass of CaCl2 required = number of moles of solute * molar mass of CaCl2

=> 0.2240 moles * 111 gm/mol

=> 24.865 grams

2) Mole fraction of solvent = 28.8/31.8 = 0.90566

Number of moles of water = Mass/molar mass = 375/18 = 20.83333 moles

Let the number of moles of compound be x

20.83333/(20.83333+x) = 0.90566

x = 2.170 moles

Molar mass of compound = 222.8 grams/2.170 = 102.66 grams

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