Question

5. Calculate the molality of C2H5OH in a water solution that is prepared by mixing 50.0 mL of C2H5OH with 108.4 mL of H2O at 20

Answer 1

Molality:

Molality is defined as the number of moles of solute dissolved in 1 kg of solvent. The molality can be expressed as follows:

                           

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First, we find the mass of C2H5OH and mass of H2O, using its mass conversion factor as:

                            

                     

From the above the data, we can find the number moles of C2H5OH and mass of C2H5OH is 39.45 g, using its molar mass as the conversion factor as:

               

Therefore, we calculate the molality as:

                           

Therefore, the molality is 7.90m and the correct option is (d) 7.90 m

Answer 2

Mass of C2H5OH = volume x concentration = 50.0 x 0.789 = 39.45 g Moles of C2H5OH = mass /molar mass of C2H5OH = 39.45/46.07 = 0.8563 mol Mass of water = volume x density - 108.4 x 1.00 108.4 g 0.1084 kg Molality moles of C2H5OH/mass of water in kg 0.8563/0.1084 = 7.90m

Answer 3

The answer is: d. 7.90 m Mass of C2H5OH = volume x density of C2H5OH -50.0 x 0.789 - 39.45 g Moles of C2H5OH mass/molar mass of C2H5OH 39.45/46.07 0.8563 mol Mass of water = volume x density of water 108.4 x 1.00 108.4 g 0.1084 kg Molality moles of C2H5OH/mass of water in kg 0.8563/0.10847.90m

Answer 4

ms= d x v 

=0.789 x 50=39.4 g

Mwt ( C2H5OH) = ( 12 X2 ) + ( 6X1 ) + ( 16X1 ) = 46 g/mol

n =ms/Mwt 

= 39.4 / 46 = 0.8 mol

 since 1ml = 1 g so mass of H2O is 100/1000 =0.1Kg

molality = 0.8 / 0.1 = 8 m