Answer: A mass of 0.518 g of phenol must be dissolved in 25.0 g of naphthalene to produce a solution that is 0.22 m in phenol.
Explanation:
Given: Mass of naphthalene = 25.0 g
Molality = 0.22 m
This means that 0.22 moles of solute is present per kg of solvent.
As 25.0 g of naphthalene is there that will be 25.0 g per 1000 g (1 kg) is equal to 0.025 kg.
Hence, moles of phenol are calculated as follows.
![Molality = \frac{moles}{mass (in kg)}\\0.22 m = \frac{moles}{0.025 kg}\\moles = 0.0055 mol](https://tex.z-dn.net/?f=Molality%20%3D%20%5Cfrac%7Bmoles%7D%7Bmass%20%28in%20kg%29%7D%5C%5C0.22%20m%20%3D%20%5Cfrac%7Bmoles%7D%7B0.025%20kg%7D%5C%5Cmoles%20%3D%200.0055%20mol)
Also, molar mass of phenol is 94.11 g/mol. This means that 1 mole of phenol contains 94.11 g.
Therefore, mass contained by 0.0055 moles of phenol is as follows.
![0.0055 mol \times 94.11 g/mol \\= 0.518 g](https://tex.z-dn.net/?f=0.0055%20mol%20%5Ctimes%2094.11%20g%2Fmol%20%5C%5C%3D%200.518%20g)
Thus, we can conclude that a mass of 0.518 g of phenol must be dissolved in 25.0 g of naphthalene to produce a solution that is 0.22 m in phenol.
Answer:
Albert Einstein is perhaps most famous for introducing the world to the equation E=mc2. In essence, he discovered that energy and mass are interchangeable, setting the stage for nuclear power—and atomic weapons. His part in the drama of nuclear war may have ended there if not for a simple refrigerator.
Explanation:
Albert Einstein is perhaps most famous for introducing the world to the equation E=mc2. In essence, he discovered that energy and mass are interchangeable, setting the stage for nuclear power—and atomic weapons. His part in the drama of nuclear war may have ended there if not for a simple refrigerator.
Answer:
Solution A: 0.00400M
Solution B: 0.00400M
Solution C: 4.00x10⁻⁵M
Explanation:
Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:
250mL / 10mL = 25 times.
That means molar concentration of sln A is:
0.100M / 25 = 0.00400M
Solution B is obtained diluting 25mL to 100mL:
100mL / 25mL = 4 times
0.00400M / 4 times = 0.00100M
And solution C is obtained diluting the solution C from 20mL to 500mL:
500mL / 20mL = 25 times
Solution C:
0.00100M / 25 times = 4.00x10⁻⁵M
C) 5p
_______________________________________________________________