Specifications for lactated Ringer’s solution, which is used for intravenous (IV) injections, are as follows to reach 100. mL of solution: \(285-315 \mathrm{mg} \mathrm{Na}^{+}\) \(14.1-17.3 \mathrm{mg} \mathrm{K}^{+}\) \(4.9-6.0 \mathrm{mg} \mathrm{Ca}^{2+}\) \(368-408 \mathrm{mg} \mathrm{Cl}^{-}\) \(231-261 \mathrm{mg}\) lactate, $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}$ a. Specify the amount of $\mathrm{NaCl}, \mathrm{KCl}, \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O},\( and \)\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\( needed to prepare \)100 . \mathrm{mL}$ lactated Ringer's solution. b. What is the range of the osmotic pressure of the solution at $37^{\circ} \mathrm{C},$ given the preceding specifications?

First, we need to find out the number of moles of each ion needed in the solution. To do this, we can use the mass of the given ion (mg) and its molar mass (g/mol). \[moles \ of \ an \ ion = \frac{mass \ of \ the \ ion}{molar \ mass \ of \ the \ ion}\]

Take the average mass of Na^+ mid-range (300 mg) and divide it by the molar mass of Na^+ (22.99 g/mol). \[moles \ of \ Na^+ = \frac{300 \ mg}{22.99 \ g/mol} = 0.0130 \ mol\]

Take the average mass of K^+ mid-range (15.7 mg) and divide it by the molar mass of K^+ (39.10 g/mol). \[moles \ of \ K^+ = \frac{15.7 \ mg}{39.10 \ g/mol} = 0.000401 \ mol\]

Take the average mass of Ca^2+ mid-range (5.45 mg) and divide it by the molar mass of Ca^2+ (40.08 g/mol). \[moles \ of \ Ca^{2+} = \frac{5.45 \ mg}{40.08 \ g/mol} = 0.000136 \ mol\]

Take the average mass of Cl^− mid-range (388 mg) and divide it by the molar mass of Cl^− (35.45 g/mol). \[moles \ of \ Cl^- = \frac{388 \ mg}{35.45 \ g/mol} = 0.01095 \ mol\]

Take the average mass of lactate ion mid-range (246 mg) and divide it by the molar mass of lactate ion (89.07 g/mol). \[moles \ of \ C_3H_5O^{-}_3 = \frac{246 \ mg}{89.07 \ g/mol} = 0.00276 \ mol\]

Using the mole ratios of each component and their molar masses, we find the mass of each substance needed to prepare the lactated Ringer's solution: NaCl: Since NaCl contains 1 Na^+ and 1 Cl^− ion, moles of NaCl = moles of Na^+ = 0.0130 mol. Mass of NaCl needed = moles of NaCl × molar mass of NaCl = 0.0130 mol × 58.44 g/mol = 0.759 g (or 759 mg) KCl: Moles of KCl = moles of K^+ = 0.000401 mol Mass of KCl needed = moles of KCl × molar mass of KCl = 0.000401 mol × 74.55 g/mol = 0.0299 g (or 29.9 mg) CaCl2·2H2O: Moles of CaCl2·2H2O = moles of Ca^2+ = 0.000136 mol Mass of CaCl2·2H2O needed = moles of CaCl2·2H2O × molar mass of CaCl2·2H2O = 0.000136 mol × 147.02 g/mol = 0.0200 g (or 20.0 mg) NaC3H5O3: Moles of NaC3H5O3 = moles of C3H5O3^− = 0.00276 mol Mass of NaC3H5O3 needed = moles of NaC3H5O3 × molar mass of NaC3H5O3 = 0.00276 mol × 112.06 g/mol = 0.309 g (or 309 mg) So, to prepare 100 mL lactated Ringer's solution, we need: - 759 mg of NaCl - 29.9 mg of KCl - 20.0 mg of CaCl2·2H2O - 309 mg of NaC3H5O3

We will use van't Hoff's law to calculate the osmotic pressure, considering the lowest and highest concentrations given. \(p_i = c_iRT\) where: - \(p_i\) = osmotic pressure of the solution - \(c_i\) = molar concentration of the ith solute in the solution - R = universal gas constant = 0.0821 atm·L/mol·K - T = temperature in Kelvin = 37 °C + 273.15 = 310.15 K To find osmotic pressures, substitute the minimum and maximum moles of each component (considering the given mass ranges) divided by 100 mL into the equation. Calculate the osmotic pressure for the lowest and highest values, sum up the partial osmotic pressures for each component, and thus find the range of the osmotic pressure.