How do you determine the percent by molarity and molality of 230 grams of sodium chloride dissolved in of 1025.0 g of water? Density of solution is 1.112g/mL.?

1 Answer
May 16, 2018

#"Molality"=3.84*mol*kg^-1#

Explanation:

#"Molality"="moles of solute"/"kilograms of solvent"#

#=((230*g)/(58.44*g*mol^-1))/(1025.0*gxx10^-3*g*kg^-1)=3.84*mol*kg^-1#.

For #"molarity"# we need to calculate the VOLUME of the resultant solution, and we have the data to do so, i.e. #rho_"solution"=1.112*g*mL^-1#

And so #"volume"="mass"/"density"=(230*g+1025*g)/(1.112*g*mL^-1)=1128.6*mL#

And so #"molarity"="moles of solute"/"volume of solution"#

#=((230*g)/(58.44*g*mol^-1))/(1128.6*mLxxxx10^-3*L*mL^-1)=3.49*mol*L^-1#

At weaker concentrations, (approx. #<0.1*mol*L^-1#) molarity and molality are equivalent....