Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.
The dichloromethane (DCM) has less density than water and also the polarity of water is much more than DCM. So the mixture of water and dichloromethane will always be a heterogeneous mixture. In the mixture dichloromethane will be always up of the water layer. The volume of the separatory funnel which contains the mixture of DCM and water must have to be more than the total volume of the liquids thus the volume of the funnel will be more than (50+50) = 100mL.
The caution have to consider during the separation are-
1. The separatory funnel have to shake well with lid and have to settle down for some times until the two liquid separated.
2. The lid should be open very slowly as the vapor pressure of DCM is more and it will float on the water.
3. After this the stopcock should be opened and slowly the water will come out first followed by DCM.
To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
There are three elements hydrogen, chlorine, and oxygen
