Question

How many kilograms are present in 6.78x1025molecules of Al2(SO4)3? (hint 1kg=103g)

Atomic masses:

Al = 27

S = 32

O = 16

A) 0.891kg

B) 38.5kg

C) Cannot be determined

D) 8.44kg

Answer 1

mol = 6.78 x 10²⁵ : 6.02 x 10²³ = 112.62

MM Al₂(SO₄)₃ = 2.27 + 3.32+12.16 =342 g/mol

mass = 112.62 x 342 = 38516.04 g = 38.5 kg

Answer 2

The total mass of 6.78*$10^{25}$ molecules of $Al_{2}(SO_{4})^3$ is (b) 38.5 Kg.

To calculate the mass, we need to first understand the values given in the question:

• 1 mole of $Al_{2}(SO_{4})^3$ contains 6.02*$10^{23}$ molecules

• 6.78*$10^{25}$/6.02*$10^{23}$ moles are in the given amount of $Al_{2}(SO_{4})^3$
• 112.62 moles of  $Al_{2}(SO_{4})^3$ is available
• Weight of 1 mole $Al_{2}(SO_{4})^3$ is 342 grams
• Hence, weight of 112.62 moles of $Al_{2}(SO_{4})^3$  = 112.62*342 grams
• The total mass is equal to 38516.04 grams, which is equal to 38.5 Kg.

Hence, the total weight of 6.78*$10^{25}$ molecules of $Al_{2}(SO_{4})^3$ is (b) 38.5 Kg.

To know more about calculation of mass from number of molecules, refer:

https://brainly.com/question/20905959

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