What is the #"molality"# of an aqueous solution of #KCl#, if #"molarity"=1*mol*L^-1#, and #rho_"solution"=1.04*g*mL^-1#?

1 Answer
Jun 3, 2017

Answer:

#"What will be molality of a KCl solution..........."# #"if MOLARITY"#
#"is 1"*mol*L^-1,# #"and"# #rho=1.04*g*mL^-1#.

Explanation:

We want the ratio.......

#"molality"="mole of solutes"/"kilograms of solvent"#.

Now in #1*L# of solution there are #1*mol# salt........, i.e. a mass of #74.55*g# salt.............. But, by specification, this SOLUTION had a mass of #1*Lxx1000*mL*L^-1xx1.04*g*mL^-1=1040*g#.

Note that I made the assumption that you meant to write #rho=1.04*g*mL^-1# NOT #1.04*g*L^-1#.

Since masses are certainly additive, given these data, we can now address the quotient...........

#"molality"="mole of solutes"/"kilograms of solvent"#

#=(1*mol_"KCl")/((1040-74.55)*gxx10^-3*kg*g^-1)=(1*mol_"KCl")/(0.9645*kg)#

#=1.037*mol*kg^-1=1.04*mol*kg^-1#....to three sigs.........