A $75 \cdot m L$ $75mL$ volume of $3.80 \cdot m o l \cdot L^{- 1}$ $3.80molL^-1$ $iron(II) sulfate$ $"iron(II) sulfate"$ is diluted to $4.0 \cdot L$ $4.0L$ . What is the new concentration?

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A $75 \cdot m L$ $75mL$ volume of $3.80 \cdot m o l \cdot L^{- 1}$ $3.80molL^-1$ $iron(II) sulfate$ $"iron(II) sulfate"$ is diluted to $4.0 \cdot L$ $4.0L$ . What is the new concentration?

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Answer 1 · 2017-02-01T18:35:24

$Molarity$ $"Molarity"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$

Explanation:

And thus we use the above equation to give us (i), the initial moles of solute:

$Moles of$ $"Moles of"$ $F e S O_{4}$ $FeSO_4$ $=$ $=$ $Volume$ $"Volume"$ $\times$ $xx$ $Molarity$ $"Molarity"$

$75 \times 10^{- 3} L \times 3.8 \cdot m o l \cdot L^{- 1}$ $75xx10^-3Lxx3.8*mol*L^-1$ $=$ $=$ $0.285 \cdot m o l$ $0.285*mol$

But this is diluted to a volume of $4.0 \cdot L$ $4.0*L$ , and so to (ii), the new concentration:

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{0.285 \cdot m o l}{4.0 \cdot L} = 0.071 \cdot m o l \cdot L^{- 1}$ $(0.285*mol)/(4.0*L)=0.071*mol*L^-1$

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A $75 \cdot m L$ $75mL$ volume of $3.80 \cdot m o l \cdot L^{- 1}$ $3.80molL^-1$ $iron(II) sulfate$ $"iron(II) sulfate"$ is diluted to $4.0 \cdot L$ $4.0L$ . What is the new concentration?

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Explanation:

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A 75⋅mL75*mL volume of 3.80⋅mol⋅L−13.80*mol*L^-1 iron(II) sulfate"iron(II) sulfate" is diluted to 4.0⋅L4.0*L. What is the new concentration?

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Explanation:

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