What mass of silver nitrate, agno3, is required to prepare 800. g of 3.50% solution of agno3

28 g of AgNo3

Explanation:

800 g of 3.5% solution of AgNo3 is to be prepared.

3.5% AgNo3 solution contains 3.5g of AgNo3 in 100 g of solution.

If 3.5 g AgNo3 is present in 100 g of solution, then how many g is needed to make 800 g of solution?

       3.5 g AgNo3100 g solution

        X g AgNo3 800 g solution

    X g AgNo3 = 3.5 x 800/100

                             = 28 g

Hence, 28 g of AgNo3 will be required to prepare 800 g of 3.5% solution of AgNo3.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%