You need a 35% alcohol solution. On hand, you have a 280 mL of a 30% alcohol mixture. You also have 75% alcohol mixture. How much of the 75% mixture will you need to add to obtain the desired solution

35 mL

Explanation:

Let the amount of 75% mixture needed be A.

The amount of each solution with their respective concentration to be added together can be expressed as:

0.75A + 0.3(280)eqn 1

The addition of the two solution must merge with the expected concentration and this can be expressed as:

0.35(A + 280)eqn 2

Eqn 1 must be equal to eqn 2, hence:

0.75A + 0.3(280) = 0.35(A + 280)

Solve for A.

0.75A + 84 = 0.35A + 98

0.40A = 14

A = 35

Hence, 35 mL of the 75% mixture will be needed.

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%

56 moles of   lieter in chemistry