Question

A) Suppose you wanted to produce 1.00L of a 3.59M solution of H2SO4

1) what is the solute?
2)what is the solvent?
3) how many grams of solute are needed to make this solution?


b) how many grams of salute are needed to make 2.50 of a 1.75M solution Ba(NO3)2?

Answer 1

1. The solute in this case is H₂SO₄, which stands for sulfuric acid.

2. The solvent is the substance in which the solute is dissolved. In this case, the solvent is not explicitly mentioned, but it is typically water (H₂O) for most aqueous solutions.

3.

Given:

Molarity (M) = 3.59 M

Volume (V) = 1.00 L

The formula to calculate the number of moles (n) of solute is:

n = M * V

Substituting the given values:

n = 3.59 mol/L * 1.00 L = 3.59 mol

To determine the mass of the solute in grams, we need to know the molar mass of H₂SO₄. The molar mass of sulfur (S) is approximately 32.07 g/mol, oxygen (O) is approximately 16.00 g/mol, and hydrogen (H) is approximately 1.01 g/mol.

The molar mass of H₂SO₄ is:

2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol

Finally, we can calculate the mass of the solute:

mass = n * molar mass

mass = 3.59 mol * 98.09 g/mol ≈ 350.96 g

Therefore, approximately 350.96 grams of H₂SO₄ are needed to make a 1.00 L solution with a molarity of 3.59 M.