The new volume : 21.85 ml
<h3>Further explanation</h3>Given
V1=25,0 ml
P1=725 mmHg
T1=298K is converted to
T2=273'K
P2=760 mmHg atm
Required
V2
Solution
Combined gas law :
Input the value :
V2=(P1.V1.T2)/(P2.T1)
V2=(725 x 25 ml x 273)/(760 x 298)
V2=21.85 ml
Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) =
where; solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6= ÷
4.6 =
4.6 × =
4.6 B = 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B =
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.