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

1 Answer
Mar 23, 2018

Answer:

#"3.3 g NaBr"#

Explanation:

A solution's mass by volume percent concentration, #"m/v %"#, tells you the number of grams of solute present in exactly #"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 #"100..0 mL"# of the solution.

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

So you can say that in order to get the equivalent of #"6.5 g"# of sodium bromide in #"100.0 mL"# of this solution, you need #"50.00 mL"# of the solution to contain

#50.00 color(red)(cancel(color(black)("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 #"3.3 g"# of sodium bromide in #"20.00 mL"# of distilled water, then add enough water to ensure that the final volume of the solution is equal ot #"50.00 mL"#.