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ollegr [7]
2 years ago
15

2 moles of a gas have a pressure of 2 atm at a temperature of 300 K. What is the volume of the gas?

Chemistry
1 answer:
qwelly [4]2 years ago
4 0
  • No of moles=2mol=n
  • Pressure=2atm=P
  • Temperature=300K=T
  • Volume=V

\\ \tt\hookrightarrow PV=nRT

\\ \tt\hookrightarrow V=\dfrac{nRT}{P}

\\ \tt\hookrightarrow V=\dfrac{2(8.314)(300)}{2}

\\ \tt\hookrightarrow V=8.314(300)

\\ \tt\hookrightarrow V=2494.2mL

\\ \tt\hookrightarrow V=24.94dm^3

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Answer:- [Na_2CO_3]=0.254M , [Na^+]=0.508 M , [CO_3^2^-]=0.254M

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL.  Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})

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sodium carbonate dissociate to give the ions as:

Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-

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[Na^+]=2(0.254M) = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

[CO_3^2^-]=0.254M


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