Question #a1f1a

2 Answers
Jul 8, 2016

Answer:

Here's what I got.

Explanation:

The idea here is that you need to use the definition of the mole fraction of this unknown solute to find a relationship between the number of moles of solute and the number of moles of solvent, which in this case is water, present in both solutions.

If you take #x# to be the number of moles of solute and #n# to be the number of moles of water present in the first solution, you can write

#(x color(red)(cancel(color(black)("moles"))))/((x + n) color(red)(cancel(color(black)("moles")))) = 0.1 -># for the undiluted solution

This will be equivalent to

#x = 0.1x + 0.1n#

#0.9x = 0.1n implies n= (0.9x)/0.1 = 9x" "color(orange)((1))#

Now, you're diluting this first solution by adding

#0.036 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "36 g"#

of water. If you take water's molar mass to be approximately equal to #"18 g mol"^(-1)#, this added mass of water will contain

#36 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18color(red)(cancel(color(black)("g")))) = "2 moles H"_2"O"#

The number of moles of solvent present in the diluted solution will now be #n + 2#. The number of moles of solute remained unchanged, since you only added water to the first solution.

You can thus say that the mole fraction of the solute in the second solution will be

#(x color(red)(cancel(color(black)("moles"))))/([x + (n+ 2)] color(red)(cancel(color(black)("moles")))) = 0.07 -># for the diluted solution

This will be equivalent to

#x = 0.07x + 0.07n + 0.14#

#0.93x= 0.07n + 0.14" "color(orange)((2))#

Now, use equation #color(orange)((1))# to write equation #color(orange)((2))# as

#0.93x = 0.07 * 9x + 0.14#

This will get you

#0.93x - 0.63x = 0.14 implies x = 0.14/0.30 = 0.4667#

This is the number of moles of solute present in both solutions.

You know that your first solution has a mass of

#0.1029 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "102.9 g"#

This solution contains #0.4667# moles of solute and, according to equation #color(orange)((1))#

#n = 9 * 0.4667 = "4.2 moles"#

of water. Use water's molar mass to calculate the mass of water

#4.2 color(red)(cancel(color(black)("moles H"_2"O"))) * "18 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "75.6 g H"_2"O"#

The mass of solute present in this solution will be

#m_"solute" = overbrace("102.9 g")^(color(blue)("mass of solution")) - overbrace("75.6 g")^(color(purple)("mass of solvent")) = "27.3 g solute"#

Since this solution contains #0.4667# moles of solute, it follows that the mass of one mole of solute, i.e. its molar mass, will be

#1 color(red)(cancel(color(black)("mole solute"))) * "27.3 g"/(0.4667color(red)(cancel(color(black)("moles solute")))) = "58.5 g"#

Therefore, the molar mass of the solute is

#"molar mass" = color(green)(|bar(ul(color(white)(a/a)color(black)("58.5 g mol"^(-1))color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one sig fig for the mole fractions of the solute in the two solutions.

This molar mass is very close to the molar mass of sodium chloride, #"NaCl"#, which is listed as

#M_("M NaCl") = "58.443 g mol"^(-1)#

so your unknown solute could very well be sodium chloride.

Jul 8, 2016

#"Let molar mass of the unidentified solute be "m g/"mol"#

#"The no. of moles of the unidentified solute be "n_s#

#"The no. of moles of water in 1st solution"=n_w#

#"Total no. of moles of water and solute in the 1st solution "#

#n_s+n_w = n#

Given

# "Mass of the solution"=0.1029kg=102.9g#

#X_1="Mole fraction of the solute in 1st solution"=0.1#

#X_2="Mole fraction of the solute in 2nd solution"=0.07#

#"Amount of water added to 1st solution" =0.036kg=36g=(36g)/(18g/"mol")=2mol#
#" where molar mass of water"=18g/(mol)#

Now applying definition of mole fraction we can write

#n_s/n=X_1 =0.1 .....(1)#

#n_s/(n+2)=X_2=0.07......(2)#

Dividing (1) by (2)

#(n+2)/n=0.1/0.07=10/7#

#=>10n=7n+14=>n =14/3#

Putting this value of n in (1)

#n_s=0.01xxn=(0.1xx14)/3=1.4/3#

Now
#n_s+n_w =n =14/3#

#=>1.4/3+n_w=14/3#

#=>n_w=14/3-1.4/3=12.6/3=4.2mol#

#" So mass of water in 1st solution"=n_wxx18=4.2xx18=75.6g#

# "Mass of solute"="Total mass of solution" -"mass of water"#

# "Mass of solute"=102.9g -75.6g=27.3g#

#"Now " n_s= 27.3/m#

#=>1.4/3=27.3/m#

#:.m = (3xx27.3)/1.4=58.5g/(mol)#

#"Hence the unidentified solute has molar mass"=58.5g/(mol)#

So the water soluble unidentified solute having molar mass 58.5g/mol may be Sodium Chloride (NaCl=23+35.5)