A particular application calls for \(\mathrm{N}_{2}(\mathrm{g})\) with a density of \(1.80 \mathrm{g} / \mathrm{L}\) at \(32^{\circ} \mathrm{C} .\) What must be the pressure of the \(\mathrm{N}_{2}(\mathrm{g})\) in millimeters of mercury? What is the molar volume under these conditions?

The molar mass of nitrogen gas can be calculated through the atomic mass of Nitrogen (\(N\)). As we know, a molecule of nitrogen gas consists of two nitrogen atoms. Therefore, the molar mass of \(N_2\) is \((2 * 14.01) g/mol\), which equals \(28.02 g/mol\).

The ideal gas law is \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. However, we know that density (\(d\)) equals mass (\(m\)) over volume (\(V\)), or \(d = m/V\). Also, we can replace mass with the product of number of moles (\(n\)) and molar mass (\(M\)), to get \(d = nM/V\). This can be rearranged to get \(n/V = d/M\). So, our ideal gas law can be rewritten as \(P = dRT/M\). The pressure \(P\) is what we are trying to find. In order to calculate this, we use the provided values: the density \(d = 1.80 g/L\), the ideal gas constant \(R = 0.0821 L atm / mol K\), and the temperature \(T = 32°C = 305K\) (converted to Kelvin, as Kelvin is the SI unit of temperature), and the calculated molar mass of nitrogen which equals \(28.02 g/mol\). As a result \(P = (1.80 g/L * 0.0821 L atm / mol K * 305 K) / 28.02 g/mol = 1.52 atm\). However, the pressure needs to be in mm Hg, and we know that \(1 atm = 760 mm Hg\), so \(P = 1.52 atm * 760 mm Hg / atm = 1155.2 mm Hg\).

The molar volume is the volume occupied by one mole of a substance at a given temperature and pressure. It's calculated by dividing the volume by the number of moles (\( n\)). Using the ideal gas law equation \( V = nRT / P \), and replacing \( n \) with the ratio of the density to molar mass ( \( d / M \) ), we get \( V = dRT / (P*M)\). Using the density \(d = 1.80 g/L\), \(R = 0.0821 L atm / mol K\), \(T = 305 K\), the calculated molar mass of nitrogen \(M = 28.02 g/mol\) and the calculated pressure value \(P = 1.52 atm\), we find the molar volume \(V = (1.80 g/L * 0.0821 L atm / mol K * 305 K) / (1.52 atm * 28.02 g/mol) = 22.4 L/mol\).