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Alla [95]
3 years ago
8

You explored the relationship between molarity, the number of moles, and solution volume. However, when you prepare a solution,

you are unable to directly measure the number of moles of solute. Instead, you can mass the solute and convert between the number of moles and mass using the molar mass of a substance as a conversion factor. The molar mass of a substance can be calculated based on its molecular formula; otherwise, it can be calculated from the mass and number of moles as follows: Molar Mass = Mass solute (in g)Moles solute In order to prepare a certain volume of solution with a desired concentration, you would need to determine the required mass of solute. What mass of CaCl2 would you need to prepare a 100. mL solution at a concentration of 0.240 M ? Express the mass in grams to three significant figures.
Chemistry
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer:

2.64gram

Explanation:

Molarity, which refers to the molar concentration, is calculated thus:

Molarity = number of moles (n) ÷ volume (L)

According to the question, volume = 100 mL, molarity = 0.240 M.

Since 1L = 1000mL

100 mL = 100/1000

= 0.1L

Hence;

0.240 = n/0.1

n = 0.1 × 0.240

n = 0.024mol

Mole = mass/molar mass

Molar mass of CaCl2 = 40 + 35.5(2)

= 40 + 71

= 111g/mol

mass = molar mass × mole

mass = 110 × 0.024

Mass = 2.64gram

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3 years ago
How many grams of AlCl3 (Molar mass =133.5 g/mol) are needed to prepare 125 mL of a 0.150 M solution?
Gnesinka [82]

Answer:

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.

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<u>0.01875 mols | 133.5 g</u>

                  <u>| 1 mol AgCl3</u>

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!!!!

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3 years ago
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