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2 years ago

1. What is the molarity of a solution prepared by dissolving 3.11 grams of NaOH in

Chemistry

1 answer:

Elena L [17]2 years ago

3 0

The molarity of a solution prepared by dissolving 3.11 grams of NaOH in

enough water to make 300 milliliters of solution is 0. 256 mol/ L

Explanation:

<h3>What is Molarity?</h3>

Molarity (M) is the amount of a substance in a given volume of solution. It is otherwise known as the number of moles of solute in a certain amount of solution.

The unit of molarity is mol/L

<h3>Formula for calculating molarity :</h3>

Molarity = number of moles of solute divided by the volume of the solution\

<h3>Parameters </h3>

Molarity = ?

Number of moles is calculated using :

Mass of the solute = 3.11 grams of NaOH

Molar mass of the solute = Na + O + H

Let's input the atomic numbers of the elements: Na (23) , O (16) , H (1)

Molar mass of NaOH = 23 + 16 + 1 = 40 g/ mol

Number of moles = 3.11 ÷ 40 = 0.077 mol

Volume of the solution is measured in Liters

Let's convert 300 milliters to liters = 300 divided by 1000 = 0.3 Liters

Molarity = 0.077 ÷ 0.3 = 0. 256 mol/ L

Therefore, the Molarity of the solution is 0. 256 mol/ L

Read more on molarity here :

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50 points I need help on this whole work sheet about converting moles

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Answer:

Explanation:

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Answer:

9.08 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of O₂ = 22.7 mol

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

O₂ : P₂O₅

5 : 2

22.7 : 2/5×22.7 = 9.08

12)

Answer:

7 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of P = 14 mol

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with P.

P : P₂O₅

4 : 2

14 : 2/4×14 = 7

13)

Answer:

76.25 mol

Given data:

Number of moles of P = 61 mol

Number of moles of O₂ react = ?

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P with O₂.

P : O₂

4 : 5

61 : 5/4×61 = 76.25

14)

Answer:

1.25 mol

Given data:

Number of moles of P₂O₅ = 0.5 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

P₂O₅ : O₂

2 : 5

0.5 : 5/2×0.5 = 1.25

15)

Answer:

20 mol

Given data:

Number of moles of P₂O₅ = 8 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

P₂O₅ : O₂

2 : 5

8 : 5/2×8 = 20

16)

Answer:

12 mol

Given data:

Number of moles of silver made = ?

Number of moles of Ag₂O = 6 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles of Ag with Ag₂O .

Ag₂O : Ag

2 : 4

6 : 4/2×6 = 12

17)

Answer:

25 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 6.25 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles of Ag with O₂ .

O₂ : Ag

1 : 4

6.25 : 4×6.25 = 25

18)

Answer:

9.8 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 2.45 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles of Ag with O₂ .

O₂ : Ag

1 : 4

2.45 : 4×2.45 = 9.8

19)

Answer:

4.4 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 2.2 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

O₂ : Ag₂O

1 : 2

2.2 : 2×2.2 = 4.4

20)

Answer:

1.5 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 0.75 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

O₂ : Ag₂O

1 : 2

0.75 : 2×0.75 = 1.5

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Explanation:

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Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

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Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

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