Carbon in the lithosphere is held in soil in the form of both organic and inorganic carbon which often as calcium carbonate. Carbon can leave the soil through soil respiration – which releases CO2, or by erosion – which can carry it into rivers or the ocean, where it then enters the hydrosphere.
The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol at equilibrium.
The volume of the bottle = 4.7 L
Mass of ethanol = 0.33 g
Temperature (T1) = -11 oC = 273-11 = 262 K
P1 = 6.65 torr
Now we will calculate the mole by applying the ideal gas equation:-
PV = nRT
Or, n = PV/RT
Where P is the pressure
T is the temperature
R is the gas constant = 0.0821 L atm mol-1K-1
V is the volume
Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-
= 0.001913 mol C2H6
Conversion of the mole to gm
Molar mass of ethanol (M) = 46.07 g/mol
Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.
Hence, the mass of ethanol present in the vapor is found to be 8.8×10⁻²g.
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Answer:- ![[Na_2CO_3]=0.254M](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D0.254M)
, ![[Na^+]=0.508 M](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D0.508%20M)
, ![[CO_3^2^-]=0.254M](https://tex.z-dn.net/?f=%5BCO_3%5E2%5E-%5D%3D0.254M)
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:
= 
So, molarity of sodium carbonate solution is 0.254 M.
sodium carbonate dissociate to give the ions as:
There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.
![[Na^+]=2(0.254M)](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D2%280.254M%29)
= 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.
