Answer to Question #151647 in Inorganic Chemistry for Eva

Question

21.3 grams of salt E (NO3) 3 are dissolved in 500 ml of solution. The concentration of nitrate ions (NO3) - in this solution is 0.6mol / l. Knowing that the E atom contains 14 neutrons in the nucleus, determine the location of element in the periodic table

Answer 1

Mass of salt = 21.3g

Volume of solution= 500ml= 0.5L

Mass concentration of salt = 21.3g/0.5L = 42.6g/L

Since nitrates are very soluble in aqueous solution, we can set the equation of dissociation below

E(NO₃)₃ ---> E³⁺ + 3NO₃^-

0.2mol/L 0.2mol/L 0.6mol/L

Since concentration of NO₃- = 0.6mol/L

Molar concentration of E(NO₃)₃ will be 0.6/3= 0.2mol/L

we can now find the molar mass of E(NO₃)₃ = mass concentration/molar concentration

= 42.6/0.2= 213g/mol

Now let's find the molar mass of E in E(NO₃)₃, let the molar mass of E be x

x + 3[14 + (16x3)] = 213

x + 186= 213

x= 213-186

x= 27g/mol

Since molar mass is equivalent to mass number

Mass number = atomic number + number of neutrons

27= n_p + 14

n_p= 27-14

n_p= 13

Therefore the atomic number of E is 13

Electronic configuration of E = 2,8,3

Therefore E belongs to group 3 and period 3 on the periodic table.

Recall that group is determined by the number of valence electrons which is 3 here, and period is determined by the number of shells which are also 3 here( occupying 2,8 and 3 electrons each).