Question

Question #e1d08

Answer 1

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 `"100.0-g"` sample and work your way from there.

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 `"1.00-L"` sample.

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` moles

`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 `"1148 g"`, and it contains `"233.76 g"` of sodium chloride, its percent concentration by mass will be

`"% 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 `"100.0-g"` sample. A solution's percent concentration by mass essentially tells you how much solute you get for every `"100 g"` of solution.

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 `"100.0-g"` sample will contain

`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 `4.000` moles of `"NaCl"` in `"1148 g"` of solution to 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 `"100.0-g"` sample's molality will be

`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.