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QveST [7]
4 years ago
10

To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

Chemistry
1 answer:
poizon [28]4 years ago
7 0
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**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 H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

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.


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Explanation:

The equilibrium reaction given in the question is as follows -

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Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

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K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

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K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

Hence ,

K' = K

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