Calculate the new molarity if 100.0 mL of water is added to 44.944.9 mL of 1.08 M sulfuric acid, H2SO4𝐻2SO4 solution. 0.335M0.335𝑀 0.485M0.485𝑀 2.41M2.41𝑀 0.007M0.007𝑀 1.45M1.45𝑀
Question
Calculate the new molarity if 100.0 mL of water is added to 44.944.9 mL of 1.08 M sulfuric acid, H2SO4𝐻2SO4 solution. 0.335M0.335𝑀 0.485M0.485𝑀 2.41M2.41𝑀 0.007M0.007𝑀 1.45M1.45𝑀
Solution
To calculate the new molarity, we need to use the formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
Given: M1 = 1.08 M (initial molarity of sulfuric acid) V1 = 44.9 mL (initial volume of sulfuric acid) V2 = 44.9 mL + 100.0 mL = 144.9 mL (final volume after adding water)
We need to find M2 (final molarity).
Rearranging the formula, we get M2 = M1V1 / V2.
Substituting the given values, we get M2 = (1.08 M * 44.9 mL) / 144.9 mL.
Calculating the above expression, we get M2 = 0.335 M.
So, the new molarity of the solution after adding 100.0 mL of water is 0.335 M.
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