# Question 43bf4

Nov 26, 2017

You bring a tall order!

#### Explanation:

We're given molarity (indirectly), and density.

So, let's find the molarity,

$\frac{230 g}{L} \cdot \frac{m o l}{58.5 g} \approx 3.93 M$

If there are $230 g$ solute and 1148g total in a liter of solution, then

(230g)/(1148g) * 100% approx 20.03%

of the solution is solute and by implication, 79.97% of the solution is solvent.

The volumetric percent of water is very close to 100%#; we're never told any other solutions are "combined".

We already deduced $3.93 m o l$ of solute are in the solution, and,

$1148 g - 230 g = 918 g \cdot \frac{m o l}{18 g} \approx 51.0 m o l$

of solvent are in the solution. Hence,

$\frac{3.93 m o l}{3.93 m o l + 51.0 m o l} \approx 0.0715$

is the mole fraction of sodium chloride, and by implication $0.928$ is the mole fraction of water.

We already deduced $918 g \cdot \frac{k g}{{10}^{3} g} = 0.918 k g$ of solvent are in the solution, and $3.93 m o l$ of solute are in the solution, hence,

$\frac{3.93 m o l}{0.918 k g} \approx 4.28 m$

Finally, the last bit is a simple unit conversion,

$230 g \cdot \frac{k g}{{10}^{3} g} \cdot \frac{2.20 l b}{k g} \approx 0.506 l b s$

of solute are distributed within the solution.