Here is the missing part of the question.
4. A distance runner is in a 4-hour race and sweats at a rate of 455 mL/hour. The sodium ion concentration in sweat is, on average, 920. mg/L. How many moles of sodium ions were lost during the race? Report your answer to three significant figures.
Answer:
Explanation:
Given that:
the time for the runner = 4 hours
the rate of sweating = 455 mL/hour
Therefore, the total sweating rate = 455 mL/hour × 4 hours
the total sweating rate = 1820 mL
The [tex]Na^+[/tex] conc. in sweat = 920 mg/L
Thus, the total number in grams of [tex]Na^+[/tex] sweat lost = [tex]\dfrac{920}{1000}\times 1820[/tex]
= 1674.4 mg
= 1.6744 grams
number of moles of [tex]Na^+[/tex] sweat lost = mass of [tex]Na^+[/tex] sweat lost/ molar mass of [tex]Na^+[/tex]
number of moles of [tex]Na^+[/tex] sweat lost = 1.6744/ 23
number of moles of [tex]Na^+[/tex] sweat lost = 0.0728 mole
5. concentration of the runner = 142 mmol/L
the average volume of blood = 5.00 L
the number of moles sodium [tex]Na^+[/tex] = concentration × volume
the number of moles sodium [tex]Na^+[/tex] = 142 mmol/L × 5.000 L
the number of moles sodium [tex]Na^+[/tex] = 710 mmol
the number of moles sodium [tex]Na^+[/tex] = 0.710 mol
recall that; the number of moles of [tex]Na^+[/tex] sweat lost = 0.0728 mol
thus, the number of remaining [tex]Na^+[/tex] = (0.71 - 0.0728) mol
the number of remaining [tex]Na^+[/tex] = 0.6372 mol
The runner ending [tex]Na^+[/tex] concentration = 0.6372 mol/ 5.00 L
The runner ending [tex]Na^+[/tex] concentration = 127440 mmol/L
