This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
There is one missing point in the question.
The formula to find an increase in boiling Temperature is :
ΔT = kb x M
ΔT = is the increase in boiling Temperature
Kb = Boiling point constant of the Solvent
M = Molarity
You did not provide the Kb. If you have it, you just have to insert it to the formula to find the ΔT.
And assuming that the other solution is water, you just have to add it up with 100 Celcius
Answer:
See below :)
Explanation:
There is an evident reason why some of the solutions Carson's has listed and observed, does conduct electricity and some that do.
A flow of electrical charge is called an electric current. Ions are atoms, or sets of atoms, that contain an electrical charge. There are two types of ions, cation or a positively charged ion containing a deficiency of electrons, and anion or a negatively charged ion which contains a surplus of electrons. When a solution conducts electricity the charge is carried within by ions that move through the solution. The larger the number of ions in the solution, the better the conductivity of the solution is. Pure water does not conduct very well because it contains very few ions, but when table salt (NaCl) is dissolved in the water, this solution does conduct well because the solution contains a more abundance of ions. The majority of the ions come from the table salt, chemically names sodium chloride. Because Sodium contains its sodium ions, and these are positive charge and chloride ions which is a negative charge, it is called an ionic substance. Not every substance is made up of ions, one such example is sugar (C12H22O11). Sugar is made up of uncharged particles also called molecules. Although sugar is a substance its molecules do not hold a charge, thus when sugar is dissolved in water, the solution does not conduct electricity, due to the lack of ions in the solution.
Therefore, depending on the ions that make up the compound, the substance would or would not conduct electricity.
A compound is where two or more elements have chemically bonded, and a mixture is where there are two or more elements are grouped together, but they are not chemically bonded.
An example of a mixture would be salt dissolved in water, and an example of a compound would be water (which is made up of 2 Hydrogen and 1 Oxygen bonded together)
tl;dr: I would go with B, basically.