Question #c028e
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"How do you calculate the Gibbs Free Energy change of a reaction?"
The molality of the #"CaCl"_2# is 1.14 mol/kg.
#"Molar mass of CaCl"_2·"2H"_2"O" = "110.98 g + 36.03 g" = "147.01 g"#
#"Mass of CaCl"_2 = 7.02 color(red)(cancel(color(black)("g CaCl"_2"·2H"_2"O"))) × "110.98 g CaCl"_2/(147.01 color(red)(cancel(color(black)("g CaCl"_2·"2H"_2"O")))) = "5.300 g CaCl"_2#
#"Mass of H"_2"O" = 7.02 color(red)(cancel(color(black)("g CaCl"_2·"2H"_2"O"))) × ("36.03 g H"_2"O")/(147.01 color(red)(cancel(color(black)("g CaCl"_2·"2H"_2"O")))) = "1.720 g H"_2"O"#
#"Total mass of water" = "40.0 g + 1.720 g" = "41.72 g"#.
So, we have a solution of 5.300 g of #"CaCl"_2# in 41.72 g of water.
#"Moles of CaCl"_2 = 5.300 color(red)(cancel(color(black)("g CaCl"_2))) × ("1 mol CaCl"_2)/(110.98 color(red)(cancel(color(black)("g CaCl"_2)))) = "0.047 75 mol CaCl"_2#
#"Molality of CaCl"_2 = "moles of solute"/"kilograms of solvent" = "0.047 75 mol"/"0.041 72 kg" = "1.14 mol/kg"#
