Question

In the laboratory you are asked to make a 0.708 m ammonium chloride solution using 315 grams of water. How many grams of ammonium chloride should you add?

Answer 1

You need to add `11.9" grams "NH_4Cl` to `315" grams of water"` to make a `0.708 " M"` solution of ammonium chloride.

Explanation:

First, let's calculate the molecular weight of ammonium chloride, `M_w^(NH_4Cl)`. For this we'll need its molecular formula, `NH_4Cl`.

`M_w^(NH_4Cl)=14.0067+4(1.00794)+35.453=53.49146`

Now we can calculate the number of grams required remembering that the density of water is 1000 gm/liter.

`315cancel(" grams water")xx(1 cancel(" liter water"))/(1000 cancel(" grams water"))xx(0.708cancel(" moles "NH_4Cl))/cancel("liter water")xx(53.49146" grams "NH_4Cl)/(cancel("mole "NH_4Cl))~~11.9" grams "NH_4Cl`