A chemist must dilute 86.3 ml of 915. mM aqueous iron(III) bromide (FeBr3) solution until the concentration falls to 628. mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume.

Question

A chemist must dilute 86.3 ml of 915. mM aqueous iron(III) bromide (FeBr3) solution until the concentration falls to 628. mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume.
Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.

GPT-4o mini · Answer

To find the final volume after dilution, we can use the dilution equation:

$C_{1} V_{1} = C_{2} V_{2}$

where:

$C_{1}$ is the initial concentration,
$V_{1}$ is the initial volume,
$C_{2}$ is the final concentration,
$V_{2}$ is the final volume.

From the problem statement:

$C_{1} = 915.0$ mM,
$V_{1} = 86.3$ mL,
$C_{2} = 628.0$ mM.

We need to find $V_{2}$ .

Substituting the values into the equation:

$(915.0, mM) \times (86.3, mL) = (628.0, mM) \times V_{2}$

Calculating the left side:

$915.0, mM \times 86.3, mL = 78974.5, mM \cdot mL$

Now, substituting this value back into the equation:

$78974.5, mM \cdot mL = 628.0, mM \times V_{2}$

Now, solve for $V_{2}$ :

$V_{2} = \frac{78974.5, mM \cdot mL}{628.0, mM} \approx 125.8, mL$

Now, convert $V_{2}$ from mL to liters:

$V_{2} = \frac{125.8, mL}{1000} = 0.1258, L$

Finally, we need to ensure our answer has the correct number of significant figures. Since our values have up to four significant figures, we will round $0.1258$ L to three significant figures, which gives:

$0.126, L$