We need to prepare 0.0021 M solution of C2SO4. But there are only 80.7 grams of the chemical available. What is the maximum volume that can be prepared?

320 L

Explanation:

We'll begin by calculating the number of mole in 80.7 g of C₂SO₄. This can be obtained as follow:

Mass of C₂SO₄ = 80.7 g

Molar mass of C₂SO₄ = (12×2) + 32 + (16×4)

= 24 + 32 + 64

= 120 g/mol

Mole of C₂SO₄ =?

Mole = mass /molar mass

Mole of C₂SO₄ = 80.7 / 120

Mole of C₂SO₄ = 0.6725 mole

Finally, we shall determine the volume. This can be obtained as follow:

Mole of C₂SO₄ = 0.6725 mole

Molarity = 0.0021 M

Volume =?

Molarity = mole / Volume

0.0021 = 0.6725 / Volume

Cross multiply

0.0021 × Volume = 0.6725

Divide both side by 0.0021

Volume = 0.6725 / 0.0021

Volume ≈ 320 L

Thus, the volume that can be prepared is approximately 320 L.

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.