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

Answer:

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#