The formula relating concentration with number of moles and volume is:

#c = frac(n)(V)#

#c# is the concentration, #n# is the number of moles, and #V# is the volume (in litres).

Now, we are given values for the volume and concentration, namely #425# #"mL"# and #3.0# #"M"#.

So let's plug these values into the equation:

#Rightarrow 3.0 = frac(n)(425 * 10^(- 3)) " " "# (volume given in litres)

#Rightarrow 3.0 = frac(n)(0.425)#

Multiplying both sides by #0.425#:

#Rightarrow 3.0 * 0.425 = frac(n)(0.425) * 0.425#

#therefore n = 1.275# #"mol"#

So, there are around #1.3# moles of solute in #425# #"mL"# of #3.0# #"M"#.

Then, let's find the mass.

Before we can do this, we need to find the molar mass #M_(m)# of #"MgCl"_(2)#:

#Rightarrow M_(m) = (24.305 + 2 * 35.453)# #"g" * "mol"^(- 1)#

#Rightarrow M_(m) = 95.211# #"g" * "mol"^(- 1)#

Now, the formula that relates number of moles to mass and molar mass is:

#n = frac(m)(M_(m))#

#n# is the number of moles, #m# is the mass, and #M_(m)# is the molar mass.

Plugging values into the equation:

#Rightarrow 1.275 = frac(m)(95.211)#

Solving for #m#:

#Rightarrow 1.275 * 95.211 = frac(m)(95.211) * 95.211#

#therefore m = 121.394025# #"g"#

So, the mass is around #121.4# grams.