Question

10) Calculate the atomic mass of element "X" if it has two naturally occurring isotopes with the following masses and natural abundances: 44.8776 g/mol 46.9443 g/mol X-45 32.88% 67.12% X-4 A) 46.34 g/mol B) 46.84 g/mol c) 44.99 g/mol D) 45.91 g/mol E) 45.60 g/mol F) 46.26 g/mol 11) Solving the Schrödinger equation does NOT allow determination of what property of an electron? A) Energy of the electron B) The exact location of the electron with 100 % certainty C) Quantum numbers of the electron D) Probability of finding an electron at a point in space E) Wavefunction of the electron F) All of the above can be determined by the Schrödinger equation 12) A sample of N2 effuses in 255 s. How long will the same size sample of Cl2 take to effuse? A) 247 s B) 645 s C) 203 s D) 388 s E) 406 s F) 155 s

Answer 1

12) According to Graham's law of diffusion , the rate of effusion or diffusion of a gas is inversely proportional to its square root of molecular mass.

Rate of effusion $\alpha$ 1/$\sqrt{}$molecular mass of the gas

=> rate of effusion  $\times$  $\sqrt{}$molecular mass of the gas = constant

r  $\times$$\sqrt{}$M = CONSTANT

=> r₁$\times$$\sqrt{}$M₁ = r₂$\times$$\sqrt{}$M₂ ------------------------------ (1)

Given the time taken for the effusion of N₂ gas = 255 s

Molecular mass of N₂ (M₁) = 28.0 g/mol

What is the time taken for the effusion of Cl₂ gas =?

Molecular mass of Cl₂ = 70.9 g/mol

Let each gas is of one mole.

since rate of effusion or diffusion = amount of gas / time taken

rate of effusion of N₂ = 1mol / 255 s

rate of effusion of Cl₂ = 1mol /time t

Substitute this values in equation (1), to get r₂ value.

(1 mol /255 s) x $\sqrt{}$28.0 g/mol =( 1 mol / time t) x  $\sqrt{}$70.9 g/mol

=> ( 1 mol / time t) = (1 mol / 255 s) x $\sqrt{}$28.0 g/mol /   $\sqrt{}$70.9 g/mol

On simplification,

(1mol / time t ) = (1 mol / 255 s) x 0.628

=> time t = 255 s / 0.628 = 406 s

Hence , time taken for the effusion of Cl₂ gas is 406 s option E.

10) Average atomic mass of an element = $\sum$mass of the isotope x its natural abundance

Substitute the given values in above equation, we get the average atomic mass of the element.

=> [44.8776 g/mol x( 32.88 /100) ]+ [46.9443 g /mol x (67.12/100)]

=> 14.7557 g/mol + 31.5090 g/mol

=> 46.2647 g/mol

=> 46.26 option F

11) Schrodinger wave equation cannot describe the exact location of electron with 100% certainity, so answer is option B,

remaining options energy levels, wave function of electron ,quantum numbers of electron and probability of finding the electron are described by it.