The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances including hexavalent chromium which is limited to 0.50 mg/L .If an industry is discharging hexavalent chromium as potassium dichromate (K2Cr2O7), what is the maximum permissible molarity of that substance?

The question asks for the maximum molarity of K₂Cr₂O₇ that can be discharge into the sewer system if the limit of hexavalent chromium (Cr(VI)) that is permmitted to discharge is 0.50 mg/L.

Molarity is defined as moles of K₂Cr₂O₇ per Liter, so we need to calculate to how many moles of K₂Cr₂O₇ corresponds the 0.50 mg of Cr(VI).

In 1 mol of K₂Cr₂O₇ there are 2 moles of Cr(VI) which are 103.99 g of Cr(VI) (2 * molar mass of Cr = 2 * 51.9961 g/mol)

So 103.99 g of Cr(VI) can be found in 1 mol of K₂Cr₂O₇

We need to calculate in how many moles of K₂Cr₂O₇ we can find 0.50 mg or 5.0x10⁻⁴ g of Cr(VI) (1 g = 1000 mg) :

x moles K₂Cr₂O₇ / 5.0x10⁻⁴ g Cr(VI) = 1 mol K₂Cr₂O₇ / 103.99 g Cr(VI)

x = 5.0x10⁻⁴ g * (1 mol K₂Cr₂O₇ / 103.99 g Cr(VI)) = 4.8x10⁻⁶ moles K₂Cr₂O₇

4.8x10⁻⁶ moles K₂Cr₂O₇ can be discharged per Liter

So the maximum permissible molarity of K₂Cr₂O₇ is 4.8x10⁻⁶ M