# 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?

Jul 13, 2016

4,980 g ${H}_{2} O$ of water is required.

#### Explanation:

$\textcolor{b l u e}{\text{We will use this equation to answer the question:}}$

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:

$11.75 \cancel{g} \times \frac{1 m o l}{236.04 \cancel{g}}$ =

$\textcolor{b r o w n}{\text{0.0498 mol HgCl}}$

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

$\text{moles solute"/"molality" = "mass of solvent (kg)}$

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

$\textcolor{b r o w n}{\text{4.98kg Solvent}}$

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

Therefore,

$4.98 \cancel{\text{kg}} \times \frac{1000 g}{1 \cancel{k} g}$ $=$

4,980 g ${H}_{2} O$