Hw Key _2 Supplemental problems

# Hw Key _2 Supplemental problems - mol Si 2 H 6 × 2 mol Si...

This preview shows page 1. Sign up to view the full content.

Chemistry 111H supplemental problems not found in the text. - Solutions These problems are adapted from Chemical Thermodynamics, An Introduction , by J. Rex Goates and J. Bevan Ott, Principles of Modern Chemistry by Oxtoby, Gillis and Nachtrieb, and Chemical Principles by Steven S. Zumdahl S1 Calculate the number of atoms of silicon (Si) in 415 cm 3 of the colorless gas disilane at 0 C and atmospheric pressure, where its density is 0.00278 g cm -3 . The molecular formula of disilane is Si 2 H 6 . To find the number of atoms, we first want to find the number of moles of Si in the sample. We can do this by using the density and the volume of the sample to find the mass of the sample, then the molar mass of Si 2 H 6 to find the moles of disilane. Once we have the moles of disilane we can find the moles of Si and the number of Si atoms using Avogadro's number. Under these conditions, 415 cm 3 of sample contains (415 cm 3 )(0.00278 g/cm 3 )=1.1537 g Si 2 H 6 1.1537 g Si 2 H 6 × mol Si 2 H 6 62.23 g Si 2 H 6 = 0.01854 mol Si 2 H 6 0.01854
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: mol Si 2 H 6 × 2 mol Si mol Si 2 H 6 = 0.03708 mol Si 0.03708 mol Si × 6.022 × 10 23 atoms / mole = 2.237 × 10 22 atoms Si S2. At its boiling point (280 °C) and at atmospheric pressure, phosphorus has a gas density of 2.7 g L-1. Under the same conditions nitrogen has a gas density of 0.62 g L-1. How many atoms of phosphorus are there in one phosphorus molecule under these conditions? According to Avogadro's hypothesis, at the same temperature and pressure, a given volume will have an equal number of molecules; therefore, the ratio of the densities will equal the ratio of the molar masses. molecular mass of P n molecular mass of N 2 = 2.7 g / L 0.62 g / L moleuclar mass of P n = 2.7 0.62 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ (28.014 g / mol ) = 122 g / mol Because the atomic mass of P is 30.97 g/atom, n must be n = 122 g / mol P n 30.97 g / atom P = 3.94 Thus, there are four P atoms in each molecule under these conditions....
View Full Document

## This note was uploaded on 02/01/2011 for the course CHEM 111 taught by Professor Owens,g during the Fall '08 term at Utah.

Ask a homework question - tutors are online