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SlyPhoenix · Mar 25, 2009, 01:45 PM

How many g of water is required to be mixed with 11.75 g of HgCl in order to make a 0.01 m solution?

Hg - 200.6

Cl - 35.45

H2O - 14.026

I have that much but I don't know hwo to set it up.

I am dreadful at Chem

Perito · Mar 25, 2009, 02:03 PM

Originally Posted by SlyPhoenix

How many g of water is required to be mixed with 11.75 g of HgCl in order to make a 0.01 m solution?

Hg - 200.6
Cl - 35.45
H2O - 14.026

The molecular weight of H2O is wrong. The atomic weight of H is 1.00794. The atomic weight of O is 15.9994. However, in this case, you don't need to know the molecular weight of water.

The molecular weight of $HgCl_2$ is $200.6 + 2(35.45) = 271.5$

OK. Now with that out of the way.

You know you need a 0.01 molar solution. You have 11.75 grams

$\frac{grams}{\frac{grams}{mole}} = mole$

${moles}\,{\frac{grams}{mole}} = grams$

$\frac{grams}{mole}$ is the molecular weight.

In your case, you have a weight that you want to convert to moles. In this case, therefore, you need to divide the weight by the molecular weight.

11.75 / 271.5 = 0.0433 moles of $HgCl_2$

To get a 0.01 molar solution, you need 0.01 moles / liter. If you dissolved it in 1 liter, you would have 0.0433 molar (moles/liter).

So,

0.0433 / X = 0.01, where X is the number of liters you need.

Solve for X and you will know the number of liters. Convert that to grams and you're finished.