Sucrose doesn't show appreciable dissociation in water, so i≈1 in this case. by extension, −7.3°C is the approximate freezing point of this solution.
<h3>What is the freezing point?</h3>
The temperature at which a liquid solidifies.
Sucrose doesn't show appreciable dissociation in water, so i≈1 in this case. by extension, −7.3°C is the approximate freezing point of this solution.
Learn more about the freezing point here:
brainly.com/question/3543605
#SPJ1
Explanation:
The given data is as follows.
Solvent 1 = benzene, Solvent 2 = water
= 2.7,
= 100 mL
= 10 mL, weight of compound = 1 g
Extract = 3
Therefore, calculate the fraction remaining as follows.
![f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5B1%20%2B%20K_%7Bp%7D%28%5Cfrac%7BV_%7BS_%7B2%7D%7D%7D%7BV_%7BS_%7B1%7D%7D%7D%29%5D%5E%7B-n%7D)
= ![[1 + 2.7(\frac{100}{10})]^{-3}](https://tex.z-dn.net/?f=%5B1%20%2B%202.7%28%5Cfrac%7B100%7D%7B10%7D%29%5D%5E%7B-3%7D)
= 
= 
Hence, weight of compound to be extracted = weight of compound - fraction remaining
= 1 - 
= 0.00001
or, = 
Thus, we can conclude that weight of compound that could be extracted is
.
The second and first one but if it isn’t 2 choices then 1