Molar Mass by Freezing Point Depression Data and Calculations Partner\'s Name Ma

Question

Molar Mass by Freezing Point Depression Data and Calculations Partner's Name Mass of test tube + "Support" +Stirrer Mass of test tube + "Support" + Stirrer +Naphthalene Data for Unknown Solid (Weigh by difference) Mass of Test Tube with Unknown Mass of Test Tube with Unknown less Sample1 Mass of Test Tube with Unknown less Sample 2 (Empty test tube) Equations of the Lines Pure Solvent 4:2o 24.404 22. 238 20.2S Trial 1 and Show your work here for solving for T: Trial 2 and Show your work here for solving for T:

Explanation / Answer

1. Molecular mass is the identity for determination of colligative properties.

2. IR spectrum determine functional group of organic molecule. And also structure of organic molecules.

3. Higher concentrations. Systematic determinate errors in preparations of solution is continue source of one error.

4 Molecule should not dissociates , assocites insoluble and sparingly soluble. Error can be associated with zero point and analytical measurements other than balance and thermameter

Answer part3

To calculate freezing point depression for 0.1 molal Nacl and Bacl2 we have 1.86 molal freezing point depression of pure water solvent with font Hoff factor 2 and 3 for NaCl And BaCl2. Freezing point depression = 1.86*0.1*2 = 0.372 K and 1.86*0.1*3 = 0.558 K respectively for NaCl and BaCl2. Now convert this kelvin scale into celcius for NaCl 273.16 -0.372 = -0.250 degree Celsius and -0.40 degree Celsius. Here this calculated values and true actual value can be accounted of molality of solutions other than 1 molal.

Answer go ahead of the part 4 first find out moles of naphthalene thenfind out molality of 1000g of caphor solvent and naphthalene solute. Number of moles of naphthalene = 0.556/128

Now 15.167g of solvent contain this solvent then 1000g of solvent contain = 0.029 moles or 0.029 molal solution concentration. The molal depression constant for camphor = 10.78/0.029 = 371 degree Celsius per mole

Answer to part 5 for each -10 degrees depression on freezing 263.15k/1.86 = 142 moles or 142*32 = 4544 g or 4.5 kg of methyl alcohol and 263.15/62 = 2.30 kg of ehelene gycol. This is preferred over methanol due to higher bp lo depressionin freezing and extensive H hydrogen bond

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