The normal boiling point for methyl hydrazine \(\left(\mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{H}_{3}\right)\) is \(87^{\circ} \mathrm{C}\). It has a vapor pressure of \(37.0 \mathrm{~mm} \mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\). What is the concentration (in \(\mathrm{g} / \mathrm{L}\) ) of methyl hydrazine if it saturates the air at \(25^{\circ} \mathrm{C}\) ?

We'll be using the following form of the Clausius-Clapeyron equation: \(\ln \frac{P_{2}}{P_{1}}=-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\) where: \(P_{1}=37.0 \text{ mm Hg}\) (vapor pressure at \(T_{1}=20^{\circ} \mathrm{C}\)) \(T_{1}=20+273.15=293.15\mathrm{K}\) \(P_{2}\) is the vapor pressure at \(T_{2}=25^{\circ} \mathrm{C}\) \(T_{2}=25+273.15=298.15\mathrm{K}\) \(\Delta H_{vap}\) is the enthalpy of vaporization, which is constant for a given substance \(R=8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) (universal gas constant) To find the enthalpy of vaporization, we first need to find out the molar mass of methyl hydrazine: \(\mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{H}_{3}\) Molar mass: \(12.01 \mathrm{g/mol} (\mathrm{C})+3(1.008 \mathrm{g/mol})(\mathrm{H})+2(14.007 \mathrm{g/mol})(\mathrm{N})=45.08 \mathrm{g/mol}\) Since methyl hydrazine boils at \(87^{\circ} \mathrm{C}\), we can assume that \(\Delta H_{vap} \approx 40.7 \text{ kJ/mol}\) which is the typical value for substances with a similar boiling point. Now we can use the Clausius-Clapeyron equation to find \(P_{2}\): \(\ln \frac{P_{2}}{37.0}=-\frac{40700}{8.314}\left(\frac{1}{298.15}-\frac{1}{293.15}\right)\) Solving for \(P_{2}\): \(P_{2}=37.0 \cdot e^{-\frac{40700}{8.314}\left(\frac{1}{298.15}-\frac{1}{293.15}\right)}=50.72 \mathrm{~mm} \mathrm{Hg}\)

Using the ideal gas law: \(PV=nRT\), we can find the concentration of methyl hydrazine. First, convert the vapor pressure from mm Hg to atm: \(P_{2}=\frac{50.72 \mathrm{~mm} \mathrm{Hg}}{760 \mathrm{~mm} \mathrm{Hg/atm}}=0.0668 \mathrm{~atm}\) Now, we will use the ideal gas law to find the number of moles of methyl hydrazine in one liter of air (\(25^{\circ} \mathrm{C}\)): \(0.0668 \mathrm{V}=n(0.0821 \cdot 298.15)\) Solving for n, assuming V = 1L: \(n=\frac{0.0668}{0.0821 \cdot 298.15}=2.72 \times 10^{-3} \text{mol}\) Finally, we will convert the number of moles to the concentration in g/L using the molar mass: Concentration = \(n \cdot \text{molar mass} =2.72 \times 10^{-3} \text{mol/L} \cdot 45.08\text{g/mol}=0.123 \mathrm{g} / \mathrm{L}\) Thus, the concentration of methyl hydrazine if it saturates the air at \(25^{\circ} \mathrm{C}\) is \(0.123 \mathrm{g} / \mathrm{L}\).