How many grams of AlCl3 (Molar mass =133.5 g/mol) are needed to prepare 125 mL of a 0.150 M solution?

A) 20.0 g AlCl3

B) 16.7g AlCl3

C) 2.50 g AlCl3

2.50 g of AlCl3

Explanation:

Goodness, stoichiometry...

So, what we need to find first is the amount of grams of AlCl3. To do this we look at the formula of molarity.

M = mols/L of solvent

So we know two parts of this formula. We have the Molarity (0.150) and the mL (125).

Now, we can't forget that we must convert 125 mL into liters so we have 0.125 L ( I forgot and had to do the entire problem again...)

So if we do the backwards equation we get:

0.150 = x/0.125

If we do math (fun ikr) we get 18.75 mols of the solution.

Now, we have to plug this wonderful number into stoichiometry

0.01875 mols | 133.5 g

                  | 1 mol AgCl3

If you are unfamiliar with what I'm doing, I'm basically going to multiply 0.01875*133.5 then divide that whole thing by 1.

So, I got 2.503125 g AlCl3

If your teacher is a stickler for significant figures there are 3 sig figs for this problem so your final answer should be

2.50 g of AlCl3

Hope you have a great day and fun with chemistry

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.