What is the final concentration of the solution produced when 225.5 mL of 0.09988 M solution of...
Question:
What is the final concentration of the solution produced when 225.5 mL of 0.09988 M solution of {eq}Na_2CO_3 {/eq} is allowed to evaporate until the solution volume is reduced to 45.00 mL?
Molarity:
The amount of solute dissolved in a solution defines the concentration of the solution. An expression of a solution's concentration is molarity, M, which can be calculated using the formula below:
{eq}\rm M = \dfrac{n}{V} {/eq}
where {eq}\rm n{/eq} is the moles of the solute and {eq}\rm V{/eq} is the volume of the solution.
Answer and Explanation: 1
We first calculate for the number of moles, n, of sodium carbonate, {eq}\rm Na_2CO_3 {/eq}, present in the given volume V of the solution, 225.5 mL or 0.2255 L, whose concentration M is 0.0988 M. Using the formula below, the moles of sodium carbonate is:
{eq}\rm n = MV = 0.09988 \frac{mol~Na_2CO_3}{L} (0.2255~L) = 0.0225229~mol~Na_2CO_3 {/eq}
This mole is also the same number of moles of sodium carbonate when the volume of the solution was reduced to 45.00 mL or 0.04500 L. We can calculate for the new molarity of the solution as follows:
{eq}\rm M = \dfrac{moles~Na_2CO_3}{volume~solution} = \dfrac{0.0225229~mol~Na_2CO_3}{0.04500~L} = \boxed{\mathbf{0.5005~M}} {/eq}
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