-2
$\begingroup$

Concentrated hydrochloric acid is $37\% \ce{HCl}$ by mass and has a density of $1.2\rm~\frac{g}{ml}$. Calculate the molarity of a solution made by diluting $125\rm~ ml$ of concentrated $\ce{HCl}$ with water to a total volume of $2\rm~ L$.

I got $194.88\rm~M$ but that's not correct. First, I found the molarity of the $37\%$ solution to be $12.18\rm~M$; then I used the $M_1V_1=M_2V_2$ formula.

Finding Molarity $$\rm\frac{100g}{1.20~\frac{g}{mL}}= 83.3~mL = 0.083~L$$ $$\rm\frac{37~g~HCl}{36.46~\frac{g}{mol}} = 1.015~mol~HCl$$

$$\rm\frac{1.015~mol}{0.083~L}= 12.18~\frac{mol}{L}$$ (This answer was marked correct on my hw)

Finding molarity of concentrated HCl $$M_1V_1=M_2V_2$$ $$M_1\times(0.125)=(12.18)(2)$$ $$M_1=194.88~\rm\frac{mol}{L}$$

$\endgroup$
2
$\begingroup$

You had the correct formula but plugged in the numbers wrong. For dilutions the formula is: $$\mathrm{M}_1 \mathrm{V}_1 = \mathrm{M}_2 \mathrm{V}_2$$ where

  • $\mathrm{M}_1$ is the original molar strength
  • $\mathrm{V}_1$ is the original volume
  • $\mathrm{M}_2$ is the diluted molar strength
  • $\mathrm{V}_1$ is the diluted volume

The formula works because $$\mathrm{M} \mathrm{V} = \mathrm{(M)(L) = (moles/L)(L) = moles}$$ so regardless of how much dilution you have the same number of moles of reagent.

So for this particular problem the final part of the solution is: $$\mathrm{(12.18M)(0.125L)}=(x)\mathrm{(2L)}$$ or $$x = \dfrac{\mathrm{(12.18M)(0.125L)}}{\mathrm{(2L)}} = \mathrm{0.761 M}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.