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Using the formula $C_1V_1 = C_2V_2$ I am supposed to calculate the molarity of dilutions.

The original solution is $\pu{0.100 M}$ at $\pu{10 mL}$. I diluted it by mixing a $\pu{7.50 mL}$ stock solution with $\pu{2.50 mL}$ water, and made another one by mixing a $\pu{5 mL}$ stock solution with $\pu{5 mL}$ water.

Intuitively, I should get a higher molarity for my first dilution since it is higher in concentration, but when I calculate using $$0.1(10)=C_2(7.5)$$ for the first dilution and $$0.1(10)=C_2(5)$$ for the second, the second dilution's molarity is higher than the first. I know I am doing something wrong but I just can't figure out what.

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  • $\begingroup$ It is a quite common mistake. Look that in first and second volume the number of molecules per unit of volume is the same, as the solution is homogeneous..; so the first equation C1 = C2, and how it is written is wrong. $\endgroup$ – user43021 Mar 8 '18 at 4:11
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You're mixing up the terms and formulae. Instead of rote learning the formulae, remember that the moles of the solute will remain the same in a solution - before and after dilution.

With that in mind, let's inspect your $\pu{7.5ml}$ stock solution. Before dilution, it had $\pu{7.5ml}\times\pu{0.1M}$ moles of the solute. After dilution, it has an unknown molarity. Let that molarity be $x ~\pu{M}$. Then, the new moles are $\pu{10ml}\times x~\pu{M}$. Equating and solving, you get $x=0.075$.

If you do that again for the second solution (I give it you as a homework), you'll get $x=0.05$. This easily verifies your own intuition that you should get "a higher molarity for the first dilution since it is higher in concentration".

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It should be $7.5\cdot0.1=(7.5+2.5)\cdot C$ then find the value of C which is the concentration of 1st solution....for the 2nd one similarly, $5\cdot 0.1=10\cdot C$ then similarly calculate concentration....

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    $\begingroup$ Nice answer! Could you elaborate on your answer (because currently OP might not understand your answer). $\endgroup$ – JSCoder says Reinstate Monica Mar 7 '18 at 16:43

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