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2. When 40.7 g of the salt HgCl is dissolved in 100 g water, the freezing point depression is observed to be 2.83 K. Based
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Answer #1

According to HOMEWORKLIB RULES we have to answer only first one problem

The freezing point depression is given by:

T = k b i

Where:

T = freezing point depression , 2.83 K

k = cryoscopic constant ;1.853 °K /m

The Kf for water is 1.853 K·kg/mol
b = molality (moles of solute per kg of solvent)
i = van't Hoff factor, r

Number of moles = amount in g / molar mass

= 40.7 g/ 271.52 g/mol

=0.1499 moles

Mass of water = 100 g or 0.100 kg

Molality = number of moles / solvent in kg

=> m =0.1499 moles /0.100 kg

= 1.499 m

Therefore calculate the i as follows:

T = k b i

2.83 K = 1.853 °K /m*1.499 m *i

i = 1.0

the value of i is 1 means it is not ionize thus it is an example of weak electrolyte which will not conduct electricity

there is no ions

Weak electrolytes:

Weak electrolytes are partially ionizes in water.

Strong electrolytes have following properties:

  1. They are ionic compounds.
  2. They are completely separates into their ions when dissolved in water.
  3. In solution form they conduct electricity due to present of ions but in solid form means neutral molecule does not conduct electricity.
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