# Question #6dd4c

##### 1 Answer

Here's how you can do that.

#### Explanation:

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

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

Now, the density of water has its **maximum value** at

#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

So, let's say that your solution has a specific gravity of **percent concentration by mass** equal to

Assuming that the density of water is given to you as

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

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

So, you know that

#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 **molar mass** of the solute, let's say *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

Since **moles** of solute, you can say that

#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 **molar mass** of

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

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