Well i suspect it to be water-- yes water btw the answers are 1B 2B 3D 4B 5B 100% Correct
Answer:
Relative volume of ether to water that should be used for the extraction = 1.205
Explanation:
The extraction/distribution coefficient of an arbitrary solvent to water for a given substance is expressed as the mass concentration of the substance in the arbitrary solvent (C₁) divided by the mass concentration of the substance in water (C₂).
K = (C₁/C₂)
Let the initial mass of the organic substance X in water be 1 g (it could be any mass basically, it is just to select a right basis, since we are basically working with percentages here).
If 94% of the organic substance X is extracted by ether in a single extraction, 0.94 g ends up in ether and 0.06 g of the organic substance X that remains in water.
Let the volume of ether required be x mL.
Let the volume of water required be y mL.
Relative volume of ether to water that should be used for the extraction = (x/y)
Mass concentration of the organic substance X in ether = (0.94/x)
Mass concentration of organic substance X in water = (0.06/y)
The distribution coefficient , Ko (Cether / C water), for an organic substance X at room temperature is 13.
13 = (0.94/x) ÷ (0.06/y)
13 = (0.94/x) × (y/0.06)
13 = (15.667y/x)
(x/y) = (15.667/13) = 1.205
Hope this Helps!!!
KOH -------------> K⁺ + OH⁻
pOH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁵ ]
pOH = 5
pH + pOH = 14
pH + 5 = 14
pH = 14 - 5
pH = 9
hope this helps!
