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.