Chemistry problem?
In 2014, a major chemical leak at a facility in West Virginia released 7500 gallons of MCHM (4-methylcyclohexylmethanol, C8H16O) into the Elk River. The density of MCHM is 0.9074 g/mL.
A) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is 7.70 feet deep, 100.0 yards wide, and 100.0 yards long. 1 gallon = 3.785 L.
B) How much farther down the river would the spill have to spread in order to achieve a "safe" MCHM concentration of 1.00×10−4M? Assume the depth and width of the river are constant and the concentration of MCHM is uniform along the length of the spill.
In 2014, a major chemical leak at a facility in West Virginia released 7500 gallons of MCHM (4-methylcyclohexylmethanol, C8H16O) into the Elk River. The density of MCHM is 0.9074 g/mL.
A) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is 7.70 feet deep, 100.0 yards wide, and 100.0 yards long. 1 gallon = 3.785 L.
B) How much farther down the river would the spill have to spread in order to achieve a "safe" MCHM concentration of 1.00×10−4M? Assume the depth and width of the river are constant and the concentration of MCHM is uniform along the length of the spill.
1 Answer
Although the scope of this problem isn't that deep, it's still fairly rigorous, especially for a beginning chemistry student.
Let's calculate the volume of water in the first part of the Elk River,
#V_"A" = 7.7"ft" * (100"yd" * (3"ft")/"yd")^2 cdot (("12 in")/("1 ft"))^3 cdot (("2.54 cm")/("1 in"))^3 cdot (("1 mL")/("1 cm"^3)) cdot (("1 L")/("1000 mL"))#
# approx 1.962*10^7"L"#
and calculate the molarity, given the preceding volume,
#"M" = (7500"gal" * (3.785"L")/"gal" * (10^3"mL")/"L" * (0.9074"g")/"mL" * "mol"/(352"g"))/V_"A"#
#approx 3.73*10^-3"M"#
For convenience later, the conversion factor from
For the next part, clearly the molar quantity won't change, but we must make some assumptions about volumetric dimensions: the width or depth must remain constant (and rectangular).
Hence, the spill would have to travel,
#10^-4 "M" = (3.73*10^-3"M" cdot 1.962*10^7"L")/V_"B"#
#=> V_"B" approx 7.32*10^8"L" * "1 ft"^3/("28.317 L") approx 2.58*10^7"ft"^3#
#ℓ = V_"B"/(underbrace("7.7 ft")_(h)*underbrace("100 yd"cdot"3 ft"/"yd")_(w)) approx 1.119*10^4"ft"#
which is an additional
