We have ammonia $25\%$. We also know $\rho_{\text{ammonia}} = \pu{0.86 \frac{g}{cm^3}}$, $\pu{25 ^\circ C}$. We want to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.

This means that there are some calculations we need to do in order to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.

I try to calculate the required mass by the following method: $\text{m}=\text{C}\cdot \text{V}\cdot M \cdot 0.25$. Is this method correct?

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Andreas Feb 24 '17 at 23:43
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    $\begingroup$ My point with dimensional analysis is to carry the units through the calculations so that you don't get twisted around... so not $$0.82\times 100=82\;\pu{g}$$ but rather $$0.82\times 100=82\;\text{g solution per 100 ml of solution }$$ not $$(0.25\times82);\pu{g}=20.5\;\pu{g}$$ but rather $$(0.25\times82)\pu{g}=20.5\;\text{g }\ce{NH3}\text{ per 100 ml of solution}$$ Other than that the calculations are fine. Good work!! $\endgroup$ – MaxW Feb 25 '17 at 0:22
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    $\begingroup$ Since you have opened a chatroom, I have cleared the comments. $\endgroup$ – orthocresol Feb 25 '17 at 0:48

$\text{Produce a solution of }3\;\text{M}\;0.5\;\pu{dm^3}\;\ce{NH3}$


Suppose 100ml

$$0.82\times 100\times 0.25=20.5\;\pu{g}\;\ce{NH3}\text{ per 100 ml of solution}$$

$$n_{ammonia}=\frac mM=\left(\frac{20.5}{17.031}\right)\pu{moles}$$ $$C_{ammonia}=\frac{n_{ammonia}}{100\times10^{-3}}\approx 12.037 \;\pu{M}.$$

To decide $V_1$, which is needed in order to prepare the solution of $0.5 \;\pu{dm^3}$ $3\;\pu{M}$ ammonia,

$$V_1=\frac{C_2V_2}{C_1}\approx0.1246 \pu{dm^3}=124.6\; \pu{ml}.$$

  • $\begingroup$ You mostly just reiterated the calculations that the OP did. :-( // Nitpicking ... Some of this isn't really right either. For instance the density should be grams per ml or kilograms per liter , not grams per liter (OP's mistake...). I'd rather that you specified that the ammonia was 25% by weight. for the equation calculating $n$ note that it is $n_{ammonia}$. $\endgroup$ – MaxW Feb 25 '17 at 17:49
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    $\begingroup$ @MaxW I'm the OP. I will add that it is 25 mass-percent. And I added my own calculations as an answer, to show that they were correct to any future reader reading this post. Should it not be g/dm_3 or must it not be g/dm_3 in order for a correct caluculation, just so I understand? I may change it, anyway. Just want to know. $\endgroup$ – Andreas Feb 25 '17 at 17:55
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    $\begingroup$ <BLUSH> Sorry. I didn't notice that you were in fact the OP. You were absolutely right to add the calcs... $\endgroup$ – MaxW Feb 25 '17 at 17:57

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