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?

1 Answer
Jul 13, 2016

4,980 g H_2O of water is required.

Explanation:

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

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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:

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Therefore,

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

4,980 g H_2O