We know that to relate solutions of with the factors of molarity and volume, we can use the equation: [tex] M_{1} V_{1} = M_{2} V_{2} [/tex]
**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M [tex] H_{2} SO_{4} [/tex] is the left side of the equation. Then we have:
[tex](18 M)(0.050 L)=(4.35M) V_{2} [/tex]
We can then solve for [tex] V_{2} [/tex]:
[tex] V_{2}= \frac{(18M)(0.05L)}{4.35M} [/tex] and [tex] V_{2} =0.21 L[/tex] or [tex]210 mL[/tex]
We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
