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STatiana [176]
3 years ago
10

The number of atoms of sulfur in 2.66 grams of sulfur

Chemistry
1 answer:
Sedbober [7]3 years ago
5 0
<h3>Answer:</h3>

5.0 × 10^21 atoms of sulfur

<h3>Explanation:</h3>

We need to know that 1 mole of an element contains atoms equivalent to the Avogadro's number.

Therefore;

1 mole of an element = 6.022 × 10^23 atoms

Thus, 1 mole of sulfur = 6.022 × 10^23 atoms

But, molar mass of sulfur = 32.065 g/mol

This means;

32.065 g contain 6.022 × 10^23 atoms

Therefore, 2.66 g of Sulfur will contain;

= (2.66 g ÷ 32.065 g/mol) ×  6.022 × 10^23 atoms

= 5.0 × 10^21 atoms

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A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),
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Answer:        

[KOH] : 1.47 M

[KOH] : 1.22 m

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Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

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