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Question #2177b

1 Answer

Jul 18, 2017

#chi_(NaOH)~=0.05#

Explanation:

Now.....#"Molality"# #=# #"Moles of solute"/"Kilograms of solvent."#

And our #NaOH(aq)# solution is #3.0*"molal"#, i.e. #3.0*mol*kg^-1#

Now #chi_(NaOH)="Moles of NaOH"/"Moles of NaOH + moles of water"#

#=(3.0*mol)/(3.0*mol+(1000*g)/(18.01*g*mol^-1))#

#=(3.0*mol)/(3.0*mol+55.52*mol)=0.0513#

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