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3 years ago

How many moles of gas would occupy a 25.0 liter when the temp is 22.0 degrees celsius and the pressure is 646 torr

Chemistry

1 answer:

Luda [366]3 years ago

3 0

Answer:

0.877 mol

Step-by-step explanation:

We can use the Ideal Gas Law to solve this problem.

pV = nRT Divide both sides by RT

n = (pV)/(RT)

Data:

p = 646 torr

V = 25.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 22.0 °C

Calculations:

(a) Convert the pressure to atmospheres

p = 646 torr × (1 atm/760 torr) = 0.8500 atm

(b) Convert the temperature to kelvins

T = (22.0 + 273.15) K = 295.15 K

(c) Calculate the number of moles

n = (0.8500 × 25.0)/(0.082 06 × 295.15)

= 0.877 mol

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Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

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Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

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Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

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Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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