Question

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

Answer 1

4,980 g `H_2O` of water is required.

Explanation:

`color(blue)("We will use this equation to answer the question:")`

We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:

`11.75 cancel gxx(1 mol)/(236.04cancelg)` =

`color (brown) ("0.0498 mol HgCl")`

Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :

`"moles solute"/"molality" = "mass of solvent (kg)"`

`(0.0498 cancel"mol") /(0.01 cancel"mol"/(kg)`

`color(brown)("4.98kg Solvent")`

Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:

Therefore,

`4.98cancel"kg"xx(1000g)/(1cancelkg)` `=`

4,980 g `H_2O`