V1 = 2.0 L
T1 = 25.0 oC = 298 K V2 = V1T2 = (2.0 L)(244 K) = 1.6 L
V2 = ? t1(298 K)
T2 = –28.9 oC = 244 K
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
Answer:
A. there is an isotope of lanthanum with an atomic mass of 138.9
Explanation:
By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.
Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.
Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.
Thus, the correct option is:
<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>
Mass divide by volume
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