Answered Essay: 4. The extracellular fluid volume varies with the size of the person. Suppose in an individual we

4. The extracellular fluid volume varies with the size of the person. Suppose in an individual we determine that the ECF is 14 L. The average [Na] in the ECF is about 143 mM. A. What is the total amount of Na in the ECF, in moles? In grams? B. Suppose this person works out and sweats 1.5 L with an average [Na] of 50 mM. During How much Na* is lost this time the urine output is 30 mL with an average [Na] of 600 mM. during the workout? C. f the person does not drink fluids at all during the workout, what will be the (Na] in the plasma at the end of the workout? Assume that all of the fluid in the sweat and urine originated from the ECF

4. The extracellular fluid volume varies with the size of the person. Suppose in an individual we determine that the ECF is 14 L. The average [Na] in the ECF is about 143 mM. A. What is the total amount of Na” in the ECF, in moles? In grams? B. Suppose this person works out and sweats 1.5 L with an average [Na] of 50 mM. During How much Na* is lost this time the urine output is 30 mL with an average [Na’] of 600 mM. during the workout? C. f the person does not drink fluids at all during the workout, what will be the (Na] in the plasma at the end of the workout? Assume that all of the fluid in the sweat and urine originated from the ECF

Expert Answer

Answer

Solution:

Total volume of ECF = 14 L

Concentration of Na = 143 mM = 0.143 M

(a). Moles of Na = Molarity * Volume

= 14 * 0.143

= 2.002 moles

Molar mass of Na = 23 g/mol

Mass of Na = 2.002 * 23

= 46.046 grams

(b). Volume of sweat = 1.5 L

Molarity of Na1 in sweat = 50 mM = 0.050 M

Moles of Na1 in sweat = 1.5 * 0.050

= 0.075 moles

Volume of urine = 30 mL = 0.030 L

Molarity of Na1 in urine = 600 mM = 0.6 M

Moles of Na1 in urine = 0.030*0.60

= 0.018 moles

Total moles of Na1 lost during workout = 0.075 + 0.018

= 0.093 moles

Mass of Na1 lost = 0.093 * 23

= 2.139 grams

(c). Moles of Na1 left in the plasma = 2.002 – 0.093

= 1.909 moles

Volume of ECF left in the body = 14 – 1.53

= 12.47 L

[Na1]  in the plasma at the end of the workout = 1.909 / 12.47

[Na1] = 0.153 M

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