General Chemistry 1311 Total: 100 pts TAMUX Chemistry Depurtment 2015 Fall Exam

Question

General Chemistry 1311 Total: 100 pts TAMUX Chemistry Depurtment 2015 Fall Exam 5 (Ch12 &Chls; 3. An alloy is compased of thr 3. An alloy is com table. What is the mole percen table. What is tlem 13) Cr and Cu. The mass thiaas sed i Fe. the percent 75.46 % 10.62 % 13.92 % Fe Cr Cu e which bas 100 grans Your workings out Assume the sampl olesans Moles Mole percentage Element Mass Mass (s) Fe Cr Cu Total moles of the elements 4. A sample 21.5 mg of the protein was dissolved in water at 15.0 °C to make 246 mL of solution and measures an osmotic pressure of 3.92 tor. What is the molar mass of this protein? Known value: R- 0.08206 atmeL/mol K Your workings out

Explanation / Answer

Solution :-

Q3

Mass percent

Fe = 75.46 %

Cr= 10.62 %

Cu = 13.92 %

If we assume mass of alloy is 100 g

Then mass of Fe= 75.46 g

Mass of Cr = 10.62 g

Mass of Cu = 13.92 g

Now lets calculate the moles of the each element

Moles = mass / molar mass

Moles of Fe= 75.46 g / 55.845 g per mol = 1.351 mol Fe

Moles of Cr= 10.62 g / 51.996 g per mol = 0.2042 mol Cr

Moles of Cu = 13.92 g / 63.546 g per mol = 0.2191 mol Cu

Now lets calculate the total moles

Total moles = 1.351 + 0.2042 +0.2191 = 1.7743

Now lets calculate the mole percent of the each element

Mol % Fe = (1.351 mol / 1.7743)*100% = 76.14 % Fe

Mol % of Cr = (0.2042 mol / 1.7743 mol )*100 % = 11.51 % Cr

Mol % of Cu = (0.2191 mol / 1.7743 mol )*100% = 12.35 % Cu

Q4) 21.5 mg

T= 15.0 C +273 = 288 K

2.46 ml = 0.00246 L

Osmotic pressure = 3.92 torr *1 atm / 760 torr = 0.005158 atm

Now lets calculate the concentration of the solution

Osmotic pressure = MRT

M= osmotic pressure / RT

    = 0.005158 atm / 0.08206 L atm per mol K * 288 K

   = 0.000218252 M

Now lets calculate the moles using the molarity and volume

Moles of protein = molarity * volume

                        = 0.000218252 mol per L * 0.00246 L

                        = 5.37*10^-7 mol

Now lets calculate the molar mass of the protein

Molar mass = mass / moles

                      = (21.5 mg * 1 g / 1000 mg) / 5.37*10^-7 mol

                      = 4.0*10^4 g /mol

So the molar mass of the protein = 4.0*10^4 g per mol

Q5)

Molar mass of H2O = (2*H)+(1*O)

                                   = (2*1.0079 g) +(1*16)

                                   = 18.0148 g per mol

Moles of H2O = 90.0 g / 18.0148 g per mol

                          = 4.996 mol

Molar mass of glucose = (6*C)+(12*H)+(6*O)

                                           = (6*12.01)+(12*1.0079)+(6*16)

                                           = 180.15 g per mol

Moles of C6H12O6 = mass / molar mass

                                  = 72.0 g / 180.15 g per mol

                                 = 0.3997 mol

Mole fraction of H2O =moles of H2O / total moles

                                       = 4.996 mol / (4.996 mol + 0.3997 mol )

                                       = 0.92592

Vapor pressure of solution = mole fraction of H2O * vapor pressure of pure H2O

                                                  = 0.92592 * 25 mmHg

                                                  =23.15 mmHg

Raoults law :- The law states that the vapor pressure of the ideal solution is proportional to the mole fraction of the solvent in the solution.

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