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.

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

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