1)a nonvolatile organic compound Z was used to make up a solution. Solution cont

Question

1)a nonvolatile organic compound Z was used to make up a solution. Solution contains 1.00g of Z dissolved in 100g of water and has vapor pressure of 754.5 mmHg at the normal boiling pt of water. Calculate the molar mass of Z in solution A. answer in g/mol

2)the osmotic pressure of 7.1x10-2M solutions of CaCl2 and urea at 25 celcius are 1.03 and 0.419 atm, respectively. Calculate the van't Hoff factor for the CaCl2 solution. i=?

3)calculate the van't Hoff factor of Na3PO4 in a 0.40 m solution whose freezing pt is -2.6 celcius

4)what is the osmotic pressure (in atm) of a 2.66 M aqueous solution of urea [(NH2)2CO] at 11.0 celcius

5) pheromones are compounds secreted by the females of many insects to attract males. one of these compounds conatins 80.78 percent C, 13.56 percent H, 5.66 percent O. A solution of 1.05 g of this pheromone in 7.93g of benzene freezes at 3.10 celcius. What are the molecular formula and molar mass of the compound? (the normal freezing pt of pure benzene is 5.50 celcius)

Molecular formula?

molar mass?

6)what are the boiling pt and freezing pt of 1.47m solution of naphthalene in benzene? (the bp and fp of benzene are 80.1 celcius and 5.5 celcius. the bp elevation constant for benzene is 2.53 celcius/m and the fp depression constant for benzene is 5.12 celcius/m

Explanation / Answer

imagine you have 100 g

80.78 g of carbon = 6.73 mol
13.56 g of H = 13.45 mol
5.66 g of O = 0.353 mol

5) pheromones are compounds secreted by the females of many insects to attract males. one of these compounds conatins 80.78 percent C, 13.56 percent H, 5.66 percent O. A solution of 1.05 g of this pheromone in 7.93g of benzene freezes at 3.10 celcius. What are the molecular formula and molar mass of the compound? (the normal freezing pt of pure benzene is 5.50 celcius)

Molecular formula?

Divide all by the lowest

6.73 / 0.353 = 19
13.45 / 0.353 = 38
0.353 / 0.353 = 1

Your empirical formula is C19H38O

The freezing point depression constant for benzene is 5.12 C kg/mol

Your depression (ha!) is 5.50 - 3.37 = 2.13 C

That means your concentration is 2.13/5.12 = 0.416 m

So in 8.5 g of solvent you have 0.416 m x 0.0085 kg = 0.00354 mol

1 g / 0.00354 mol = 282 g/mol

Your empirical formula weighs 12(19) + 38 + 16 = 282

So you have one empirical formula per molecular formula and your molar mass is 282.

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