# Question de361

Apr 10, 2016

Here's how you can do that.

#### Explanation:

In order to convert the given concentration of sodium cations, ${\text{Na}}^{+}$, from milligrams per liter, ${\text{mg L}}^{- 1}$, to millimoles per liter, ${\text{mmol L}}^{- 1}$, you need to use the molar mass of elemental sodium.

Since you already know the concentration of the cations per liter, you can focus on converting milligrams to millimoles.

A sodium cation has the same molar mass as elemental sodium, $\text{Na}$

${M}_{M} = {\text{22.989770 g mol}}^{- 1}$

As you know, you can go from grams to milligrams by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 g" = 10^3"mg}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You can use the same conversion factor to go from moles to millimoles

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 mol" = 10^3"mmol}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that grams per mole,${\text{g mol}}^{- 1}$, will be equivalent to milligrams per millimole, ${\text{mg mmol}}^{- 1}$, since

[color(red)(cancel(color(black)("g")))/color(blue)(cancel(color(black)("mol"))) * (10^3"mg")/(1color(red)(cancel(color(black)("g")))) * (1color(blue)(cancel(color(black)("mol"))))/(10^3"mmol") = color(purple)(cancel(color(black)(10^3)))/color(purple)(cancel(color(black)(10^3)))"mg"/"mmol" = "1 mg mmol"^(-1)]#

This means that the molar mass of the sodium cations can be expressed as

${\text{22.989770 g mol"^(-1) = "22.989770 mg mmol}}^{- 1}$

You can use this value as a conversion factor to go from milligrams per liter to mmoles per liter

$8.034846159 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mg")))/"L" * "1 mmol Na"^(+)/(22.989770color(red)(cancel(color(black)("mg")))) = color(green)(|bar(ul(color(white)(a/a)"0.3494965870 mmol L}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to ten sig figs, since that's how many sig figs you have for the concentration of the sodium cations.