Question #6dd4c

For starters, the specific gravity of a substance is simply the ratio between its density and the density of a reference substance, which more often than not is water at #4^@"C"#.

#"SG" = rho_"substance"/rho_ ("H"_ 2"O at 4"^@"C")#

Now, the density of water has its maximum value at #4^@"C"#

#rho_ ("H"_ 2"O at 4"^@"C") = "0.99997 g mL"^(-1)#

but if the problem doesn't specify this, you can approximate it with #"1.0 g mL"^(-1)#.

So, let's say that your solution has a specific gravity of #"SG"# and a percent concentration by mass equal to #x# #%#, which means that #"100 g"# of this solution will contain #x# #"g"# of solute.

Assuming that the density of water is given to you as #"1.0 g mL"^(-1)#, you can say that the density of the solution will be

#rho_"solution" = "SG" * "1.0 g mL"^(-1)#

#rho_"solution" = "SG" quad "g mL"^(-1)#

So, you know that #"1 mL"# of this solution has a mass of #("SG")# #"g"#, so pick a #"100-g"# sample and calculate its volume.

#100 color(red)(cancel(color(black)("g solution"))) * "1 mL"/(("SG") color(red)(cancel(color(black)("g solution")))) = (100/"SG") quad "mL"#

This sample will contain #x# #"g"# of the solute, so use the molar mass of the solute, let's say #M_M# #"g mol"^(-1)#, to calculate the number of moles of solute present in the sample.

#x color(red)(cancel(color(black)("g solute"))) * "1 mole solute"/(M_M color(red)(cancel(color(black)("g solute")))) = (x/M_M) quad "moles solute"#

Now, in order to find the molarity of the solution, you need to determine how many moles of solute are present in #"1 L" = 10^3 quad "mL"# of the solution.

Since #(100/"SG")# #"mL"# of this solution contain #(x/M_M)# moles of solute, you can say that #10^3# #"mL"# of the solution will contain

#10^3 color(red)(cancel(color(black)("mL solution"))) * ( (x/M_M) quad "moles solute")/((100/"SG") color(red)(cancel(color(black)("mL solution")))) = ((10 * x * "SG")/M_M) quad "moles solute"#

You can thus say that the molarity of a solution that has a specific gravity of #"SG"#--with the density of water equal to #"1.0 g mL"^(-1)#--a percent concentration by mass equal to #x%#, and a solute that has a molar mass of #M_M# #"g mol"^(-1)# is equal to

#"molarity" = ((10 * x * "SG")/M_M) quad "moles L"^(-1)#

This solution contains #((10 * x * "SG")/M_M)# moles of solute for every #"1 L" = 10^3 quad "mL"# of the solution.