Question

in order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00 kg of water?

Answer 1

If you are talking about a 0.523 molar KI solution.  You have to find the volume of the solution which can usually be assumed to just be that of water which would be 2 L in this case since 1mL=1g of water.  Then you multiply 0.523M by 2L to get 1.046 moles of KI.  After you get the number of moles you multiply that by its molar mass to get the grams.  
(1.046 moles KI)x(165 grams/mole KI)= 172.59g KI

If you are talking about a 0.523 molal KI solution you have to multiply 0.523 molal by 2Kg to get 1.046 moles of KI.  like before you have to multiply that by its molar mass to get grams.

The units for molarity is moles/L and the units of molality is moles(solute)/Kg(solvent)

Even though in this question the molarity and molatity appear to be the same thing, they are really not.  Do not get the two confused with each other because if you have not seen already, they are different ways of measuring concentration and have their own uses.

Answer 2

Answer:

$m_{KI}=173.74gKI$

Explanation:

Hello,

In this case, one recalls the formula of molality due to the 0.523 m solution and find the moles of potassium iodide as shown below:

$m=\frac{n_{KI}}{m_{solvent}} \\n_{KI}=m*m_{solvent}=0.523\frac{molKI}{kgH_2O}*2.00kgH_2O=1.046molKI$

Now, by using the potassium iodide's atomic mass, one computes the grams:

$m_{KI}=1.046molKI*\frac{166gKI}{1molKI} =173.74gKI$

Best regards.