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Chapter 22: Problem 54

When \(1.645 \mathrm{~g}\) of white phosphorus are dissolved in \(75.5 \mathrm{~g}\) of \(\mathrm{CS}_{2},\) the solution boils at \(46.709^{\circ} \mathrm{C}\) whereas pure \(\mathrm{CS}_{2}\) boils at \(46.300^{\circ} \mathrm{C}\). The molal boiling-point elevation constant for \(\mathrm{CS}_{2}\) is \(2.34^{\circ} \mathrm{C} / \mathrm{m}\). Calculate the molar mass of white phosphorus and give the molecular formula.

Short Answer

Expert verified
The molar mass of white Phosphorus is approximately 106.4 g/mol and its molecular formula is \( P_4 \).

Step by step solution

01

Calculate the molality of the solution

Using the formula of boiling point elevation, \( \Delta T = k_b \cdot m \), where \( \Delta T \) is the change in temperature, \( k_b \) is the boiling point elevation constant and \( m \) is the molality. From the problem, \( \Delta T = 46.709^{\circ}C - 46.300^{\circ}C = 0.409^{\circ}C \), and \( k_b = 2.34^{\circ}C/m \). Therefore, the molality \( m = \Delta T / k_b = 0.409^{\circ}C /2.34^{\circ}C/m = 0.175 mol/kg \). Note that mol/kg is the same as molal.
02

Calculate the molar mass of white phosphorus

Using the definition of molality, which is the number of moles of solute per kilogram of solvent, the molar mass (M) of white phosphorus can be calculated using \( M = m / w \), where \( w \) is a mass of white phosphorus in kilograms. From the problem, \( w = 1.645g = 0.001645kg \). So, the molar mass of white phosphorus is \( M = 0.175 mol/kg / 0.001645kg = 106.4 g/mol \).
03

Determine the molecular formula of white phosphorus

Phosphorus (P) has an atomic weight of approximately \( 31.0 g/mol \). Therefore, the molecular formula is determined by dividing the molar mass of the white phosphorus by the atomic weight of phosphorus. This yields \( 106.4 g/mol / 31.0 g/mol \approx 3.43 \). Rounding to the nearest whole number, the molecular formula becomes \( P_4 \).

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