Answer:
615ml
Explanation:
Vol(ml) = (gms solute)/[(Molar Answer => Conc)(formula Wt solute))] = (100g)/{(1.4M)(116g/mol)} = 0.615L = 615ml.
Explanation:
The given data is as follows.
Partition coefficient, K = 6.2
Volume of phase 1,
= 74.0 mL
Volume of phase 2,
= 17.0 mL
So, after one extraction fraction of solute remaining is given as follows.
q = 
After 3 times extraction, fraction of S remaining is as follows.
q = ![[\frac{V_{1}}{V_{1} + KV_{2}}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BV_%7B1%7D%7D%7BV_%7B1%7D%20%2B%20KV_%7B2%7D%7D%5D%5E%7B3%7D)
Putting the given values into the above formula as follows.
q = ![[\frac{V_{1}}{V_{1} + KV_{2}}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BV_%7B1%7D%7D%7BV_%7B1%7D%20%2B%20KV_%7B2%7D%7D%5D%5E%7B3%7D)
= ![[\frac{74.0 ml}{74.0 ml + 6.2 \times 17.0 ml}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B74.0%20ml%7D%7B74.0%20ml%20%2B%206.2%20%5Ctimes%2017.0%20ml%7D%5D%5E%7B3%7D)
= ![[\frac{74.0 ml}{179.4 ml}]^{3}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B74.0%20ml%7D%7B179.4%20ml%7D%5D%5E%7B3%7D)
= 0.0699
Thus, we can conclude that the fraction of S remaining in the aqueous phase is 0.0699.
Answer:
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Answer:
Well unless you have your testicales removed but I don't really know