What is the final concentration of a $25.7 \cdot m L$ $25.7mL$ volume of $K m n O_{4} (a q)$ $KmnO_4(aq)$ that is diluted to a $50.00 \cdot m L$ $50.00mL$ volume?

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What is the final concentration of a $25.7 \cdot m L$ $25.7mL$ volume of $K m n O_{4} (a q)$ $KmnO_4(aq)$ that is diluted to a $50.00 \cdot m L$ $50.00mL$ volume?

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Answer 1 · 2016-09-21T19:21:56

Approx. $1.56 \cdot m o l \cdot L^{- 1}$ $1.56*mol*L^-1$ , as you have diluted the starting solution by half.

Explanation:

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution (L)}$ $"Moles of solute"/"Volume of solution (L)"$ , and thus $Concentration$ $"Concentration"$ has the units $m o l \cdot L^{- 1}$ $mol*L^-1$ . With this relationship we can solve most problems of dilution.

We started with $25.7 \cdot m L$ $25.7*mL$ of a $3.12 \cdot m o l \cdot L^{- 1}$ $3.12*mol*L^-1$ $K M n O_{4}$ $KMnO_4$ solution.

$Moles of K M n O_{4}$ $"Moles of "KMnO_4$ $=$ $=$ $25.7 \times 10^{- 3} \cdot L \times 3.12 \cdot m o l \cdot L^{- 1}$ $25.7xx10^-3*Lxx3.12*mol*L^-1$ $=$ $=$ $? ? \cdot m o l$ $??*mol$ .

This molar quantity was then diluted to a $50.00 \cdot m L$ $50.00*mL$ volume.

$Final concentration$ $"Final concentration"$ $=$ $=$ $(25.7xx10^-3*cancelLxx3.12*mol*cancel(L^-1))/(50.00xx10^-3*L)$ $=$ $??mol*L^-1$

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What is the final concentration of a $25.7 \cdot m L$ $25.7mL$ volume of $K m n O_{4} (a q)$ $KmnO_4(aq)$ that is diluted to a $50.00 \cdot m L$ $50.00mL$ volume?

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What is the final concentration of a 25.7*mL25.7⋅mL25.7*mL volume of KmnO_4(aq)KmnO4(aq)KmnO_4(aq) that is diluted to a 50.00*mL50.00⋅mL50.00*mL volume?

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What is the final concentration of a $25.7 \cdot m L$ $25.7mL$ volume of $K m n O_{4} (a q)$ $KmnO_4(aq)$ that is diluted to a $50.00 \cdot m L$ $50.00mL$ volume?