Question

calculate the volume in milliliters of a 0.24m aluminum sulfate solution that contains 200. mmol of aluminum sulfate (al₂(so₄)₃). round your answer to 2 significant digits.

Explanation:

Step1: Recall the formula for molarity

Molarity ($M$) is defined as $M=\frac{n}{V}$, where $n$ is the number of moles (in this case, $n = 200\space mmol=0.200\space mol$) and $V$ is the volume in liters. We can rearrange the formula to solve for $V$: $V=\frac{n}{M}$.

Step2: Substitute the given values into the formula

Given $M = 0.24\space M$ and $n=0.200\space mol$, we substitute these values into the formula: $V=\frac{0.200\space mol}{0.24\space mol/L}$.

Step3: Calculate the volume in liters

$V=\frac{0.200}{0.24}\space L=\frac{200}{240}\space L=\frac{5}{6}\space L\approx0.833\space L$.

Step4: Convert liters to milliliters

Since $1\space L = 1000\space mL$, we multiply the volume in liters by 1000: $V = 0.833\space L\times1000\space mL/L = 833\space mL$. Rounding to 2 significant digits, we get $V = 830\space mL$.

Answer:

$830\space mL$