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Chapter 5: Problem 86

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) and trichlorosilane \(\left(\mathrm{SiHCl}_{3}\right)\) are both starting materials for the production of electronics-grade silicon. Calculate the densities of pure \(\mathrm{SiCl}_{4}\) and pure \(\mathrm{SiHCl}_{4}\) vapor at $85^{\circ} \mathrm{C}$ and 635 torr.

Short Answer

Expert verified
The densities of pure Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) and pure trichlorosilane \(\left(\mathrm{SiHCl}_{3}\right)\) vapor at \(85^{\circ}\mathrm{C}\) and 635 torr are \(4.87\,\text{g/L}\) and \(3.90\,\text{g/L}\), respectively.

Step by step solution

01

Calculate the molar mass of both substances

First, we need to find the molar mass of Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) and trichlorosilane \(\left(\mathrm{SiHCl}_{3}\right)\). Remember that to find the molar mass, simply add the atomic masses of the individual elements in the chemical formula (given in g/mol). Silicon tetrachloride: Molar mass of Si = 28.09 g/mol Molar mass of Cl = 35.45 g/mol So, molar mass of \(\mathrm{SiCl}_{4}= 28.09 + 4\times 35.45 = 169.39\,\text{g/mol}\) Trichlorosilane: Molar mass of Si = 28.09 g/mol Molar mass of H = 1.01 g/mol Molar mass of Cl = 35.45 g/mol So, molar mass of \(\mathrm{SiHCl}_{3}= 28.09 + 1.01 + 3\times 35.45 = 135.44\,\text{g/mol}\)
02

Convert the given pressure and temperature to appropriate units

We need to convert the given pressure and temperature to the SI units before using the Ideal Gas Law to calculate the density of vapor. Pressure = 635 torr, convert to atmospheres (atm) using the conversion factor: \(1\, \text{atm} = 760\, \text{torr}\) So, \(P = \frac{635\, \text{torr}}{760\, \text{torr/atm}} = 0.8368\,\text{atm}\) Temperature = \(85^{\circ}\mathrm{C}\), convert to Kelvin (K) using the conversion: \(T_{\text{K}} = T_{\text{C}} + 273.15\) So, \(T = 85 + 273.15 = 358.15\,\text{K}\)
03

Calculate the volume of one mole of gas using Ideal Gas Law

Using the Ideal Gas Law, \(PV = nRT\), we can find the volume of one mole of gas (i.e., n = 1 mol) by solving for V. For both \(\mathrm{SiCl}_{4}\) and \(\mathrm{SiHCl}_{3}\), we have: \(V = \frac{nRT}{P} = \frac{1\,mol\times 0.0821\, \mathrm{L\, atm/mol\, K} \times 358.15\, \mathrm{K}}{0.8368\, \mathrm{atm}}\) \$V_{\mathrm{SiCl}_{4}} = \frac{1\,mol\times 0.0821\, \mathrm{L\, atm/mol\, K} \times 358.15\, \mathrm{K}}{0.8368\, \mathrm{atm}} = 34.78\, \mathrm{L/mol}\$ \$V_{\mathrm{SiHCl}_{ださい

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