Boron has only to naturally occurring isotopes. the mass of boron-10 is 10.01294 amu and the mass of bornon-11 is 11.00931 amu. calculate the relative abundances of the two isotopes

X=percent as a fraction of boron 10 
y=percent as a fraction of boron 11 

The percentages as fractions add to equal one 

x+y=1 

x=1-y 

10.01294x+11.00931y=10.811 

substitute in the value of x from equation 1 in to equation 2 

10.01294(1-y)+11.00931y=10.811 

multiply out and solve for y 

10.01294-10.01294y+11.00931y=10.811 

0.99637y=0.79806 

y=0.800967512= approx 80.1 % boron 11 isotope 

100%-80.1%=19.9% boron 10 isotope 

e) Na₃PO₄

Explanation:

a) BaSO₄

1 mole of BaSO ₄ = 6.022×10²³ molecules

1 mole of BaSO ₄ contain 1 atom of Ba one atom of S and four atoms of O.

Total number of atoms = 6 atoms

6.022×10²³ × 6 atoms = 36.132 ×10²³ atoms

b) NaNO₂

1 mole of NaNO₂= 6.022×10²³ molecules

1 mole of NaNO₂ contain 1 atom of Na one atom of N and two atoms of O.

Total number of atoms = 4 atoms

6.022×10²³ × 4 atoms = 24.088 ×10²³ atoms

c)KMnO₄

1 mole of KMnO₄ = 6.022×10²³ molecules

1 mole of KMnO₄ contain 1 atom of K one atom of Mn and four atoms of O.

Total number of atoms = 6 atoms

6.022×10²³ × 6 atoms = 36.132 ×10²³ atoms

d) KCl

1 mole of KCl = 6.022×10²³ molecules

1 mole of KCl contain 1 atom of K one atom of Cl.

Total number of atoms = 2 atoms

6.022×10²³ × 2 atoms = 12.044 ×10²³ atoms

e) Na₃PO₄

1 mole of Na₃PO₄ = 6.022×10²³ molecules

1 mole of Na₃PO₄ contain 3 atom of Na one atom of P and four atoms of O.

Total number of atoms = 8 atoms

6.022×10²³ × 8 atoms = 48.176×10²³ atoms