How to calculate the volume of HNO3 needed to produce a specific concentration? [closed]








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> 
> How much $\pu{cm^3}$ of $\ce{HNO3}$, $\omega\_{\ce{(HNO3)}} = \pu{63}\%$ with density $\ce{HNO3}$ of $\pu{1.4 g cm-3}$ do we need to create $\pu{500 cm^3}$ of solution whose $c = \pu{0.2 mol dm-3}$?
> 
> 
> 


I tried to solve it and got following answer:


\begin{align}
V\_\ce{(HNO3)} &= ? \\
ω\_\ce{(HNO3)} &= 63\% \\
\rho\_\ce{(HNO3)} &= \pu{1.4 g cm-3} \\
V\_\mathrm{(solution)} &= \pu{500 cm3} \\
c &= \pu{0.2 mol dm-3} \\
c &= \frac{n\_\ce{(HNO3)}}{V\_\mathrm{(solution)}} \\
n\_\ce{(HNO3)} &= c \times V\_\mathrm{(solution)} \\
n\_\ce{(HNO3)} &= \pu{0.2 mol dm-3} \times \pu{0.5dm^3}\\
n\_\ce{(HNO3)} &= \pu{0.1 mol}\\
m\_\ce{(HNO3)} &= \pu{6.3 g}\\
m\_\mathrm{(solution)} &= \frac{m\_\ce{(HNO3)} \times 100\%}{\omega\_\ce{(HNO3)}}\\
m\_\mathrm{(solution)} &= \frac{\pu{6.3 g} \times 100\%}{63\%}\\
m\_\mathrm{(solution)} &= \pu{10 g}\\
\rho\_\ce{(HNO3)} &= \frac{m\_\ce{(HNO3)}}{V\_\ce{(HNO3)}}\\
V\_\ce{(HNO3)} &= \frac{m\_\ce{(HNO3)}}{\rho\_\ce{(HNO3)}}\\
V\_\ce{(HNO3)} &= \frac{\pu{6.3 g}}{\pu{1.4 g cm-3}}\\
V\_\ce{(HNO3)} &= \pu{4.5 cm^3}
\end{align}


Calculating mass of solution here wasn't needed. Also, my friends did the same exercise and they got answer: $\pu{7.14 cm^3}$


This is new lesson that we are taking and I quite don't understand it. Our teacher wrote these as formulas:


\begin{align}
\omega\_\mathrm{solute} &= \frac{m\_\mathrm{solute}}{m\_\mathrm{solution}}\\
m\_\mathrm{solution} &= m\_\mathrm{solute} + m\_\mathrm{solvent}\\
\rho &= \frac{m\_\mathrm{solution}}{V}\\
c\_\mathrm{supstance} &= \frac{n\_\mathrm{supstance}}{V\_\mathrm{solution}}
\end{align}


Are these formulas correct? If I made mistake, please point it out. I don't understand this part at all. 




Frankly I think that you did so many individual calculations that you lost track of what you were doing. Obviously the amount (moles or grams) of concentrated acid used equals the amount of acid in the dilute solution. Let's use the subscript $\mathrm{i}$ for the initial concentrated solution and the subscript $\mathrm{f}$ for the subscript for the final solution.


$$c\_\mathrm{i}\times V\_\mathrm{i} = c\_\mathrm{f}\times V\_\mathrm{f}\tag{1}$$


Rearranging 


$$V\_\mathrm{i} = \frac{c\_\mathrm{f}\times V\_\mathrm{f}}{c\_\mathrm{i}}\tag{2}$$


Now $V\_\mathrm{f}$ has units of $\pu{cm^3}$, which is fine for a volume measurement. But $c\_\mathrm{f}$ has units of $\pu{mol \cdot dm^3}$ but the concentrated acid is given as $63\%$ by weight. So we need to convert the concentration acid to $\pu{mol \cdot dm^3}$ so that the units cancel. 


$$c\_\mathrm{i} = \frac{0.63 \times \pu{1.4 g/cm^3} \times \pu{1000 cm^3/dm^3}}{\pu{63 g/mol}} = \pu{14.0 mol/dm^3}\tag{3}$$


Now going back to equation 2


$$V\_\mathrm{i} 
= \frac{c\_\mathrm{f}\times V\_\mathrm{f}}{c\_\mathrm{i}} 
= \frac{\pu{0.2 mol/dm^3}\times \pu{500 cm^3}}{\pu{14.0 mol/dm^3}} 
= \pu{7.1 cm^3}\tag{4}$$


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