Question

lets say a sample was collected from a local water treatment plant. the 100 ml sample contained 0.000450 grams of sodium fluoride.
what would the concentration of this sample be?
m
the epa considers concentrations that are greater than 0.000095 m to be cause for concern. would this sample exceed that limit?
yes
no
molarity (m) = \frac{moles of solute}{liters of solution}
molarity is the amount of moles of a substance dissolved in a liter of solution (usually a water - based solution).
1 liter is equivalent to 1000 ml.

Explanation:

Step1: Calculate moles of sodium fluoride

Use formula \(n=\frac{m}{M}\), where \(m = 0.000450\space g\) and \(M=47\space g/mol\).
\(n=\frac{0.000450}{47}\space mol\approx9.5745\times10^{-6}\space mol\)

Step2: Convert volume to liters

Given \(V = 100\space mL\). Since \(1\space L=1000\space mL\), then \(V=\frac{100}{1000}\space L = 0.1\space L\)

Step3: Calculate molarity

Use formula \(M=\frac{n}{V}\). Substitute \(n = 9.5745\times10^{-6}\space mol\) and \(V = 0.1\space L\).
\(M=\frac{9.5745\times 10^{-6}}{0.1}\space M=9.5745\times10^{-5}\space M\approx0.000096\space M\)

Step4: Compare with EPA limit

EPA limit is \(0.000095\space M\). Since \(0.000096> 0.000095\)

Answer:

\(0.000096\space M\); Yes