# Question #a1f1a

##### 2 Answers

#### 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 **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 theundiluted 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 **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 thediluted 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 **moles** of solute and, according to equation

#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 **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,

#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)**