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QUESTION

# Nitric acid is usually purchased in concentrated form with a 70.3% HNO3 concentration by mass and a density of 1.41 g/mL.How much of the concentrated stock solution in milliliters should you use to make 1.5L of 0.500M HNO3?

##"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 .