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Dennis_Churaev [7]
3 years ago
13

What is the molar mass of 1-butene if 5.38 × 1016 molecules of 1-butene weigh 5.00 μg?

Chemistry
1 answer:
xenn [34]3 years ago
4 0
<h3>Answer:</h3>

             Molar Mass =  56 g.mol⁻¹

<h3>Explanation:</h3>

Data Given:

                   Mass  =  5.00 μg  =  5.0 × 10⁻⁶ g

                   Number of Molecules  =  5.38 × 10¹⁶ Molecules

Step 1: Calculate Moles of 1-Butene:

                      As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Particles ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  5.38 × 10¹⁶ Molecules ÷ 6.022 × 10²³

                          Number of Moles  =  8.93 × 10⁻⁸ Moles

Step 2: Calculate Molar Mass of 1-Butene:

As,

                          Mole  =  Mass ÷ M.Mass

Solving for M.Mass,

                          M.Mass  =  Mass ÷ Mole

Putting values,

                          M.Mass  =  5.0 × 10⁻⁶ g ÷ 8.93 × 10⁻⁸ mol

                          M.Mass =  55.99 g.mol⁻¹ ≈ 56 g.mol⁻¹

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6 0
3 years ago
Read 2 more answers
A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

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