##”47.8 mL”##
A solution’s is defined as the mass of the divided by the total mass of the solution and multiplied by 100.
Nitric acid’s concentration by mass is given to be ##70.3%##; this means that for every ##”100 g”## of solution we get ##”70.3 g”## of ##”HNO”_3##.
Let’s assume we have a ##”1-L”## bottle of stock ##”HNO”_3## solution (we could assume any volume, the answer must always be the same – in this case, 1L makes the calculations easier).
Knowing that the solution’s is ##”1.41 g/mL”##, which is equal to
##1.41″g”/color(red)(cancel(color(black)(“mL”))) * (1000 color(red)(cancel(color(black)(“mL”))))/”1 L” = “1410 g/L”##
we can determine the stock solution’s mass
##”density” = “mass”/”volume”##
so you get
##”mass” = “density” * “volume”##
##m_”sol” = 1410″g”/color(red)(cancel(color(black)(“L”))) * 1color(red)(cancel(color(black)(“L”))) = “1410 g”##.
Out of this quantity, ##70.3%## represents ##HNO_3##, so nitric acid’s mass will be
##1410 color(red)(cancel(color(black)(“g solution”))) * (“70.3 g HNO”””_3)/(100color(red)(cancel(color(black)(“g solution”)))) = “991.2 g HNO”””_3##
Knowing that nitric acid’s molar mass is ##”63 g/mol”##, we determine the number of moles to be
##991.2 color(red)(cancel(color(black)(“g”))) * “1 mole”/(63 color(red)(cancel(color(black)(“g”)))) = “15.7 moles”##
Thus, its is
##C = n/V = “15.7 moles”/”1 L” = “15.7 M”##
We can now apply calculations to determine what the requested volume will be
##C_1V_1 = C_2V_2##
##V_1 = C_2/C_1 * V_2 = (0.5 color(red)(cancel(color(black)(“M”))))/(15.7color(red)(cancel(color(black)(“M”)))) * 1.5 * 10^3″mL” = color(green)(“47.8 mL”)##.
I’ll leave the answer rounded to three .
