# Question #e1d08

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you need to pick a *volume sample* of this solution and determine its percent concentration by mass by using its density.

#color(blue)("% w/w" = "mass of solute"/"mass of solution" xx 100)#

Once you know its percent concentration by mass, you can apply it to the

Since you're dealing with molarity, which as you know is defined as moles of solute per **liters** of solution, you can make the calculations easier by picking a

This sample will contain

#color(blue)(c = n/V implies n = c * V)#

#n = "4.000 M" * "1.00 L" = "4.000 moles NaCl"#

Now use the solution's density to determine its *mass*

#1.00 color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace( "1.148 g"/(1color(red)(cancel(color(black)("mL")))))^(color(blue)("density")) = "1148 g"#

Use sodium chloride's molar mass to determine how many grams would contain

#4.000 color(red)(cancel(color(black)("moles"))) * "58.44 g NaCl"/(1color(red)(cancel(color(black)("mole")))) = "233.76 g NaCl"#

So, if the solution has a total mass of

#"% w/w" = (233.76 color(red)(cancel(color(black)("g"))))/(1148color(red)(cancel(color(black)("g")))) xx 100 = color(green)("20.36%")#

Now focus on the **for every**

In your case, you'd get

#100.0color(red)(cancel(color(black)("g solution"))) * "20.36 g NaCl"/(100color(red)(cancel(color(black)("g solution")))) = "20.36 g NaCl"#

This of course means that the

#m_"water" = m_"solution" - m_"solute"#

#m_"water" = "100.0 g" - "20.36 g" = color(green)("79.64 g H"_2"O")#

A solution's molality is defined as the number of moles of solute per **kilogram of solvent**.

#color(blue)("molality" = "moles of solute"/"kilograms of solution")#

Now, how many *moles* of sodium chloride would this sample contain?

Use the fact that you get

#100.0 color(red)(cancel(color(black)("g solution"))) * "4.000 moles NaCl"/(1148color(red)(cancel(color(black)("g solution")))) = "0.3484 moles NaCl"#

This means that the

#b = "0.3484 moles"/(79.64 * 10^(-3)"kg") = color(green)("4.375 molal")#

Finally, the solution's concentration in *grams per liter* will be

#"233.76 g"/"1.00 L" = color(green)("233.8 g/L")#

All the answers are rounded to four sig figs.

**SIDE NOTE** *As practice, you can try calculating the concentration in grams per liter of the 100.0-g sample*.

*It must come out to be exactly the same as the concentrations in grams per liter of the initial, 1.00-L sample, plus/minus a small error from rounding I did along the way.*