A good first step is writing the amount in terms of ml.
19.2 gallons = 72.68 L = 72680 ml
that would mean it weighs 0.749*720680g = 54437.32ml = 54.437 L
hope that helps :)
Answer: C) 0.020 m
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

where,
n = moles of solute
= weight of solvent in g
Mole fraction of
is =
i.e.
moles of
is present in 1 mole of solution.
Moles of solute
= 
moles of solvent (water) = 1 -
= 0.99
weight of solvent =
Molality =
Thus approximate molality of
in this solution is 0.020 m
Answer:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.00064M)
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
![K=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Which can be written in terms of x, according to the ICE table:

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!