A solution of #"1.25 g"# of a non-electrolyte in #"20 g"# of water freezes at #"271.94 K"#. If #K_f = "1.86 K"cdot"m"^(-1)#, then the molecular weight of the solute is?

1 Answer
Junaid Mirza · Truong-Son N.
Nov 11, 2017

#"96.15 g/mol"#

Explanation:

Pure water freezes at #"273.15 K"#.

Freezing point depression (#DeltaT_f#) for nonelectrolytic solutions is given by:

#DeltaT_f = T_f - T_f^"*" = -K_f m#

where #K_f# is the freezing point depression constant and #m# is the molality in #"mols solute/kg solvent"# for a freezing point #T_f#. #"*"# indicates pure substance.

Substituting values in, we get:

#"271.94 K" - "273.15 K" = -("1.86 K")/"molal" × "m"#

#=> m = "0.65 molal"#

Again, molality of a solution is given by

#"Molality" = "Moles of solute" / "Mass of solvent (in kg)"#

From above,

#"mols solute" = m cdot "Mass of solvent (in kg)"#

#= "0.65 mols solute"/cancel"kg solvent" cdot cancel"1 kg"/(1000 cancel"g") xx 20 cancel"g"#

#=# #"0.013 mols solute"#

#color(blue)("Molar Mass") = "g of solute" / "mols of solute"#

#= "1.25 g" / "0.013 mols" = color(blue)("96.15 g/mol")#

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