15.On an unusually warm day in February, Monica (a burger-flipper by default) is

Question

15.On an unusually warm day in February, Monica (a burger-flipper by default) is sweating profusely over a hot grill. The temperature in the fast food kitchen is 1080 F or 42.20 C and the relative humidity is 88%. During her one morning break, she takes a drink of some ice cold distilled water from a plastic cup. Before Monica takes a drink, she drips 6 drops of sweat into the cup. Given: the concentration of sodium in her sweat is 70 mmol/L, the concentration of potassium in sweat is 8.5 mmol/L, the pH of her the sweat is 6.0, her sweat production rate is 2.1 L/hr, a drop of sweat is 0.1 mL, the concentration of sodium in the plastic cup after her sweat is added is 2.68 mg/L, and that Monica would rather be studying physiology than “sweating her brains out”. How much fluid is in the plastic cup before she takes a sip?

Explanation / Answer

There's a lot of unneeded informatino here! The question is asking what the volume of liquid in the cup is.

We know that 0.6 mL of sweat has been added (6 drops at 0.1 mL each), and that the final sodium concentration is 0.25 mg/L.

We know the initial concentration of sodium in the sweat is 70 mmol/L. So, this is just a dilution problem!

For a dilution problem, we use:

c1 * V1 = cf * Vf.

C1 is the intial concentration of sodium (=70 mmol/L).
V1 is the initial volume (0.6 mL)
cf is the final concentration (2.68 mg/L)
Vf is the final volume (what we want).

The catch here is that c1 and cf must be in the same units.

The MW of sodium is 23 g/mol, or 23 mg/mmol.
So, 70 mmol/L = (23 mg/mmol * 70 mmol/L) = 1610 mg/L.

Now we can just plug in:

1610 mg/L * 0.6 mL = 2.68 mg/L * Vf

Vf = (1610mg/L) * (0.6 mL) / (2.68 mg/L)

The units cancel out, so we are left with Vf = (1610 * 0.6 / 2.68) mL.

So the volume of fluid in the cup before she drinks is 371 mL

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