Question

Find the value of x if hydrated salt A₂​SO₄​⋅xH₂​O undergoes a 45% loss in mass on heating and becomes anhydrous. The atomic mass of A is 7.

Answer 1

To find the number of water molecules (x) in the hydrated salt A2SO4·xH2O, we need to solve for x using the given mass loss percentage and atomic mass of element A.

Step-by-step explanation:

We need to find the value of x for the compound A2SO4·xH2O, which experiences a 45% mass loss upon heating to become anhydrous. Suppose the initial mass of the hydrated salt is m grams. After heating, the anhydrous form weighs 0.55m grams, given the 45% mass loss.

Let's calculate the molar mass of the anhydrous part, A2SO4. With the atomic mass of A being 7, we find the molar mass as follows: (2×7) + (1×32) + (4×16) = 14 + 32 + 64 = 110 grams per mole (g/mol).

If x moles of water are lost, that corresponds to x × 18 (molar mass of water) grams. The mass of the hydrated form can be expressed as m = 110 + x×18 grams and the anhydrous form as 0.55m = 110 grams. Using the mass loss percentage:

110 + x×18 = × × 0.55 (110 + x×18)

Solving this equation gives us the value of x, which is the number of water molecules originally present in the hydrated salt.