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elixir [45]
3 years ago
14

To make a 2.0-mole solution, how many moles of solute must be dissolved in 0.50 liters of solution?

Chemistry
1 answer:
Flauer [41]3 years ago
3 0

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. The concentration of the solution is calculated as follows:

 Molarity = 2.0 mole / L solution

<span>2.0 mole / L solution ( 0.50 Liters ) = 1 mole solute</span>

<span>The correct answer is the third option. One mole of solute needed to make 0.50 liters of 2M solution.</span>

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A gas occupies a volume of 650.0 mL when the pressure is 3.50 atm. What will the new volume be if the pressure is reduced to 1.6
Gemiola [76]

The new volume will be 1379 mL.

Explanation:

As per Boyle's law, the product of initial volume and initial pressure of any gas molecule is equal to the product of final volume and final pressure of those molecules.

So here the initial volume is 650 ml and the initial pressure is 3.50 atm. As the temperature is said to be constant, then this system will be obeying Boyle's law. So, the final pressure is given as 1.65 atm. As there is a reduction in the pressure, the volume of the gas is tend to get expanded.

P_{1} V_{1} = P_{2} V_{2}

So, (650*10^{-3}*3.50)=(1.65*V_{2})

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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
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Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

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

T = temperature

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

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

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