Final answer:
To prepare 8.00 liters of Ringer's lactate, 25.08 grams of sodium lactate are required, based on the concentration of lactate in Ringer's solution and the molecular weight of sodium lactate.
Step-by-step explanation:
To determine how many grams of sodium lactate are needed to prepare 8.00 liters of Ringer's lactate, we need to consider both the concentration of sodium lactate in Ringer's lactate and the volume of solution we want to prepare. Given that Ringer's lactate contains about 28 mEq/L lactate, we have to first convert mEq to moles, recognizing that mEq is a measurement of the amount of substance based on charge, and for sodium lactate (which has a 1+ charge for Na+), 1 mEq is equivalent to 1 mmol. We must then use the molecular weight of sodium lactate to convert moles to grams.
Since there are 28 mEq of lactate per liter of Ringer's solution and we need 8 liters, we calculate the total mEq as 28 mEq/L x 8 L = 224 mEq.
Now, assuming the molecular weight of sodium lactate (NaC3H5O3) is approximately 112 g/mol, we convert mEq to moles (remember that 1 Eq is 1 mol for monovalent ions like Na+), and hence 224 mEq is equivalent to 224 mmol or 0.224 mol.
Finally, we calculate the mass required using the molecular weight: 0.224 mol x 112 g/mol = 25.08 grams of sodium lactate.