What is the molarity of a stock solution if 60 mL were used to make 150 mlof a .5M solution?
1 answer:
The molarity of the stock solution is 1.25 M.
<u>Explanation:</u>
We have to find the molarity of the stock solution using the law of volumetric analysis as,
V1M1 = V2M2
V1 = 150 ml
M1 = 0.5 M
V2 = 60 ml
M2 = ?
The above equation can be rearranged to get M2 as,
M2 =
Plugin the values as,
M2 =
= 1.25 M
So the molarity of the stock solution is 1.25 M.
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Explanation:
It is known that equation for ideal gas is as follows.
PV = nRT
The given data is as follows.
Pressure, P = 1500 psia, Temperature, T = = 104 + 460 = 564 R
Volume, V = 2.4 cubic ft, R = 10.73
Also, we know that number of moles is equal to mass divided by molar mass of the gas.
n =
m =
=
= 9.54 lb
Hence, molecular weight of the gas is 9.54 lb.
- We will calculate the density as follows.
d =
=
= 3.975
- Now, calculate the specific gravity of the gas as follows.
Specific gravity relative to air =
=
= 51.96
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