A sample of drinking water severely contaminated with chloroform , CHCl3 ,is supposed to be carcinogenic in nature . The level of combination was 15 ppm (by mass) .What is the molality of chloroform in the water sample ?

2 Answers
May 17, 2016

Answer:

#"15 ppm" ~=# #(15xx10^-3*g)/(119.37*g*mol^-1)xx1/(1*kg)# #=# #??*mol*kg^-1#

Explanation:

#"1 ppm"# #=# #1*mg*L^-1#, by definition.

If there is a #"15 ppm"# solution of chloroform in water, this represents, #(15xx10^-3*g)/(119.37*g*mol^-1)xx1/(1*kg)# #=# #??*mol*kg^-1#, with respect to chloroform.

You might object to this treatment in that the chloroform solution will have a slightly different density than pure water, but it will have a negligible effect. This is quite probably concentrated enough to taste.

May 17, 2016

Answer:

The molality of chloroform is #1.2 × 10^"-4"color(white)(l) "mol/kg"#.

Explanation:

The formula for molality is

#color(blue)(|bar(ul(color(white)(a/a) "Molality" = "moles of solute"/"kilograms of solute"color(white)(a/a)|)))" "#

Assume that you have 1 L of water. Then the mass of water is 1000 g.

#"Mass of CHCl"_3 = 1000 color(red)(cancel(color(black)("g water"))) × "15 g CHCl"_3/(10^6 color(red)(cancel(color(black)("g water")))) = "0.015 g CHCl"_3#

#"Moles of CHCl"_3 = 0.015 color(red)(cancel(color(black)("g CHCl"_3))) × "1 mol CHCl"_3/(119.38 color(red)(cancel(color(black)("g CHCl"_3))) ) = 1.2 × 10^"-4"color(white)(l) "mol CHCl"_3#

#"Molality" = (1.2 × 10^"-4" color(white)(l)"mol")/"1 kg" = 1.2 × 10^"-4" color(white)(l)"mol/kg"#