# What mass of "NaBr" is needed to prepare "50.00 mL" of a 6.5% (m/v) solution in distilled water?

Mar 23, 2018

$\text{3.3 g NaBr}$

#### Explanation:

A solution's mass by volume percent concentration, $\text{m/v %}$, tells you the number of grams of solute present in exactly $\text{100.0 mL}$ of the solution.

So all you need to know in order to figure out a solution's mass by volume percent concentration is the number of grams of solute present for every $\text{100..0 mL}$ of the solution.

In your case, you know that you're dealing with a 6.5% $\text{m/v}$ solution, which means that every $\text{100.0 mL}$ of this solution contain $\text{6.5 g}$ of sodium bromide, the solute.

So you can say that in order to get the equivalent of $\text{6.5 g}$ of sodium bromide in $\text{100.0 mL}$ of this solution, you need $\text{50.00 mL}$ of the solution to contain

$50.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL solution"))) * "6.5 g NaBr"/(100.0color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("3.3 g NaBr}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass by volume percent concentration.

To make this solution, you would dissolve $\text{3.3 g}$ of sodium bromide in $\text{20.00 mL}$ of distilled water, then add enough water to ensure that the final volume of the solution is equal ot $\text{50.00 mL}$.