A solid mixture contains \(\mathrm{MgCl}_{2}\) and \(\mathrm{NaCl}\). When \(0.5000 \mathrm{g}\) of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at \(25.0^{\circ} \mathrm{C}\) is observed to be 0.3950 atm. What is the mass percent of \(\mathrm{MgCl}_{2}\) in the solid? (Assume ideal behavior for the solution.)

Given the osmotic pressure (\(\Pi\)) is 0.3950 atm and the temperature (\(T\)) is \(25.0^{\circ} \mathrm{C}\), find the molar concentration ([C]) using the osmotic pressure equation: \(\Pi = [C]RT\) Rearrange the equation to find [C]: \([C] = \frac{\Pi}{RT}\) To find R, the ideal gas constant, we use: R = 0.0821 \(L\cdot atm \cdot mol^{-1}K^{-1}\) To convert the temperature to Kelvins, we add 273.15 to the Celsius temperature: T(K) = TH(°C) + 273.15 Plug in the values and calculate: \([C] = \frac{0.3950 \,atm}{(0.0821 \, L\cdot atm \cdot mol^{-1}K^{-1})(25+273.15)K}\)

Let x and y represent the mole fractions of \(\mathrm{MgCl}_{2}\) and \(\mathrm{NaCl}\) respectively, where x+y=1. Since the ions produced by each salt are different, we need to consider that each salt contributes to the total molarity with a different ratio: \([C] = x([C]_{MgCl_{2}}) + y([C]_{NaCl})\) Since we have been given the total mass(new_mass) as 0.5000g, we can represent the mass of each salt as follows: mass of \(MgCl_{2}\) = xm mass of \(NaCl\) = ym Where xm + ym = 0.5000g.

From the previous steps, we have the equations: 1. [C] = x([C]_{MgCl_{2}}) + y([C]_{NaCl}) 2. xm + ym = 0.5000g With \(\mathrm{MgCl}_{2}\) and \(\mathrm{NaCl}\) having molar masses of 95.211 g/mol and 58.44 g/mol, respectively, we can write the equation in terms of mass: \[0.5000 = x\left(\frac{95.211}{y(58.44)}\right)\] Solve the equation for x: \[x =\frac{0.5000 \cdot y(58.44)}{(95.211)}\] Substitute the value x into the mole fractions equation \(x + y = 1\) and solve for y: \[y = \frac{95.211}{153.651}\] Then, find the value of x: \[x = 1 - y = \frac{58.44}{153.651}\] With now the x value, we determine the mass percent of \(\mathrm{MgCl}_{2}\) in the solid mixture: Mass percent of \(\mathrm{MgCl}_{2}\) = \(\frac{x}{x + y} \times 100\%\) Mass percent of \(\mathrm{MgCl}_{2}\) = \(\frac{\frac{58.44}{153.651}}{\frac{153.651}{153.651}} \times 100\%\) Therefore, the mass percent of \(\mathrm{MgCl}_{2}\) in the solid mixture is approximately 38.00%.