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sp2606 [1]
3 years ago
5

Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume

of 0.175 L at a temperature of 25.0◦C is held at a pressure of 1.000 atm. What gas is this most likely?
Neon
Chlorine
Fluorine
Oxygen
Krypton
Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

             mass = 0.508 g,               Volume = 0.175 L

             Temperature = (25 + 273) K = 298 K,       P = 1 atm

As per the ideal gas law, PV = nRT.

where,  n = no. of moles = \frac{mass}{\text{molar mass}}

Hence, putting all the given values into the ideal gas equation as follows.

               PV = \frac{mass}{\text{molar mass}} \times RT            

           1 atm \times 0.175 L = \frac{0.508 g}{\text{molar mass}} \times 0.0821 L atm/ K mol \times 298 K  

                            = 71.02 g

As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of Cl_{2} is 70.8 g/mol or 71 g/mol (approx).

Thus, we can conclude that the gas is most likely chlorine.

You might be interested in
How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm?
meriva
Hello!

Data:

P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)

Apply the data to the Clapeyron equation (ideal gas equation), see:
P*V = n*R*T

1*18.5 = n*0.082*300

18.5 = 24.6n

24.6n = 18.5

n =  \dfrac{18.5}{24.6}

\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark

Note: If the feedback is to be considered, the closest response is 0.751 mol Nacl

_________________
_________________

I hope this helps. =)


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