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
.
<span>The "second" is the SI base unit of time.</span>
That is correct c
Explanation
Answer:
1.373 mol H₂O
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
24.75 g H₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
<u />
= 1.37347 mol H₂O
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig fig rules and round.</em>
1.37347 mol H₂O ≈ 1.373 mol H₂O
The molality of the solution is obtained as 0.63 m.
<h3>What is the freezing point?</h3>
The freezing point is the temperature at which the liquid is converted into solid.
We know that;
ΔT = 3.5° C
K = 1.86° C/m
i = 3
m = ?
Thus;
ΔT = K m i
m = ΔT/K i
m = 3.5° C/ 1.86° C/m * 3
m = 0.63 m
Learn more about freezing point:brainly.com/question/3121416
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