Question #a1f1a
2 Answers
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 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
#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
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
#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
#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
#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,
#M_("M NaCl") = "58.443 g mol"^(-1)#
so your unknown solute could very well be sodium chloride.
Given
Now applying definition of mole fraction we can write
Dividing (1) by (2)
Putting this value of n in (1)
Now
So the water soluble unidentified solute having molar mass 58.5g/mol may be Sodium Chloride (NaCl=23+35.5)