Question

1. An aqueous solution of zinc(II) fluoride is prepared by dissolving 5.61 g of zinc(II) fluoride in 7.58×10² g of water. The density of the solution is 1.42 g mL^-1.

Determine the molarity (in mol/L) of zinc(II) fluoride in the solution.

2. An element consists of two isotopes with the following natural abundances:

	Natural Abundance (%)	Atomic Mass (amu)
isotope 1	47.079	107.440
isotope 2	52.921	?

If the average atomic mass of the element is 107.964 amu, calculate the isotopic mass of the second isotope.

(Report your answer to three decimal places.)

3. Name the orbital (1s, 2s, 2p, 3s, 3p, 3d etc...) given by each set of quantum numbers below.



{n, l, m_l, m_s} = {7, 3, -1, -1/2}

{n, l, m_l, m_s} = {5, 2, 0, 1/2}

{n, l, m_l, m_s} = {7, 1, 1, -1/2}

{n, l, m_l, m_s} = {3, 2, 1, 1/2}

{n, l, m_l, m_s} = {6, 1, 1, 1/2}

{n, l, m_l, m_s} = {7, 1, -1, 1/2}

Answer 1

1.

Molarity =

W₂ is mass of Zinc fluoride = 5.61 g.

M₂ is molar mass of Zinc fluoride (ZnF₂) = 103.4 g/mol.

V = volume of solution (mL) = ( mass of solution /density)

= (5.61+ 7.58*10²)/1.42

= 537.75 mL.

So, molarity = (5.61*1000)/(103.4*537.75)

= 0.1008 (mol/L) .

2.

Average atomic mass =

Or,

107.964 = (47.079*107.440 + 52.921*x)/100

Or, 10796.4 = 5058.167 + 52.921x

Or ,52.921x = 5738.233

Or, x = 108.430 amu.

Hence atomic mass of second isotope 108.430 amu.

3.

l = 3 means f orbital. l = 2 means d orbital ,l = 1 means p orbital.

Hence first one is 7f.

Second is 5d.

Third is 7p.

Fourth is 3d

Fifth is 6p.

Sixth is 7p.