# Question #39909

Mar 14, 2015

You'd need $5.00 \cdot {10}^{3} \text{mL}$ of sodium hydroxide $\text{0.125 M}$-solution to provide that many grams of sodium hydroxide to your reaction.

You know that concentration is defined as the number of moles of solute, in your case sodium hydroxide, or $N a O H$, per liter of solution.

Mathematically, this is equivalent to

$C = \frac{n}{V}$

Because you need to solve for volume, the above equation will be more useful in this form

$V = \frac{n}{C}$

Since you know the concentration of the solution, all you need to know in order to figure out your required volume is the number of moles present in 25.0 g , which you can get using sodium hydroxide's molar mass

$\text{25.0 g NaOH" * "1 mole NaOH"/"40.0 g" = "0.625 moles NaOH}$

Plug this into the equation and you'll get

$V = \frac{n}{C} = \text{0.625 moles"/"0.125 mol/L" = "5.00 L}$

Expressed in mililiters, the volume is equal to

$\text{5.00 L" * "1000 mL"/"1 L" = "5,000 mL}$

Rounded to three sig figs, the answer will be

$V = 5.00 \cdot {10}^{3} \text{mL}$