Question

5.If the same distance runner started with a blood sodium ion concentration of 142 mmol/L and drank water equivalent to the volume of sweat lost during the race, what is the runner’s ending sodium ion concentration in mmol/L? Average blood volume is 5.00 L. Report your answer to three significant figures. Hint: If the runner drank as much fluid as was lost, what is the runner’s total blood volume at the end of the race? How many moles of sodium ions were in the runner’s blood at the end of the race?

Answer 1

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:

Step-by-step 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
`Na^+` conc. in sweat = 920 mg/L

Thus, the total number in grams of
`Na^+` sweat lost =
`(920)/(1000)* 1820`

= 1674.4 mg

= 1.6744 grams

number of moles of
`Na^+` sweat lost = mass of
`Na^+` sweat lost/ molar mass of
`Na^+`

number of moles of
`Na^+` sweat lost = 1.6744/ 23

number of moles of
`Na^+` 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
`Na^+` = concentration × volume

the number of moles sodium
`Na^+` = 142 mmol/L × 5.000 L

the number of moles sodium
`Na^+` = 710 mmol

the number of moles sodium
`Na^+` = 0.710 mol

recall that; the number of moles of
`Na^+` sweat lost = 0.0728 mol

thus, the number of remaining
`Na^+` = (0.71 - 0.0728) mol

the number of remaining
`Na^+` = 0.6372 mol

The runner ending
`Na^+` concentration = 0.6372 mol/ 5.00 L

The runner ending
`Na^+` concentration = 127440 mmol/L