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Sulfuric acid (\(H_2SO_4\)) has a molar mass of 98.1 g/mol. How many oxygen atoms are found in 75.0 g of \(H_2SO_4\)?

A. \(1.15 \times 10^{23}\)

B. \(1.84 \times 10^{24}\)

C. \(4.61 \times 10^{23}\)

D. \(7.87 \times 10^{23}\)

Answer :

Mass present in 1 gram of mole substance is called molar mass. Molar mass is calculated by: dividing the mass of the compound in grams by the number of moles present in the substances.

The correct answer is:

option B. 1.84 × 10²⁴

Given,

  • The mass of sulfuric acid =75.0 g

  • The molar mass of sulfuric acid = 98.1 g/mol

1.The moles of sulfuric acid can be calculated by

M = m ÷ n

where M is molar mass, m is mass and n moles.

moles of sulfuric acid = mass ÷ molar mass

= 75 g ÷ 98.1 g/mol

=0.7645 moles

Hence moles of sulfuric acid present is 0.7645 moles.

2. 1 mole of sulfuric acid = 4 oxygen atoms.

Then, 0.7645 moles of sulfuric will contain: 0.7645 × 4 = 3.0581 atoms of oxygen

3. Oxygen atoms present in 75.0 g of H₂SO₄ will be calculated by multiplying it with Avogadro's constant:

= 3.0581 x 6.022× 10²³

= 1.84x10²⁴

Therefore, 1.84 × 10²⁴ oxygen atoms will be found in 75.0 g of H₂SO₄.

To learn more about moles follow the given link:

https://brainly.com/question/20486415

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Rewritten by : Barada

Answer:

B

Explanation:

first find the no. Of moles of H2SO4

= 75/98.1

= 0.7645

Next step

In on mole of H2SO4 the no.of oxygen atom is 4.

1 H2SO4 : 4 O

0.7645 : x x= 0.7645x4= 3.0581

next multiply avogadros constant

3.0581 x 6.022× 10^23 =1.84x10^24