# Question fa843

Jul 29, 2017

$5.95$ $\text{g NaOH}$

#### Explanation:

We're asked to find the mass, in $\text{g}$, of $\text{NaOH}$ in $248.0$ $\text{mL}$ of a $0.600 M$ $\text{NaOH soln}$.

To do this, we can use the molarity equation to calculate the number of moles of $\text{NaOH}$:

$\text{molarity" = "mol solute"/"L soln}$

The molarity is given as color(green)(0.600M, and the volume is $248.0$ $\text{mL}$.

Since this volume must be in liters, we have

248.0cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = color(red)(0.2480 color(red)("L"

Let's find the moles of $\text{NaOH}$:

"mol NaOH" = ("molarity")("L soln") = (color(green)(0.600M))(color(red)(0.2480color(white)(l)"L"))

= color(purple)(0.1488 color(purple)("mol NaOH"

Lastly, we'll use the molar mass of sodium hydroxide ($40.00$ $\text{g/mol}$) to find the number of grams:

color(purple)(0.1488)cancel(color(purple)("mol NaOH"))((40.00color(white)(l)"g NaOH")/(1cancel("mol NaOH"))) = color(blue)(ul(5.95color(white)(l)"g NaOH"#