What is the #"molality"# of an aqueous solution of #KCl#, if #"molarity"=1molL^-1#, and #rho_"solution"=1.04gmL^-1#?

Answer 1

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

The ratio is what we're looking for.

#"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#.

Given these facts, we can now address the quotient since masses are undoubtedly additive.

#"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.........

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Answer 2

Wyatt Atkins

The molality of the aqueous solution of KCl is approximately 0.958 mol/kg.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

What is the #"molality"# of an aqueous solution of #KCl#, if #"molarity"=1molL^-1#, and #rho_"solution"=1.04gmL^-1#?