# Question 6dd4c

Jan 18, 2018

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 ${4}^{\circ} \text{C}$.

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

Now, the density of water has its maximum value at ${4}^{\circ} \text{C}$

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 ${\text{1.0 g mL}}^{- 1}$.

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

Assuming that the density of water is given to you as ${\text{1.0 g mL}}^{- 1}$, you can say that the density of the solution will be

${\rho}_{\text{solution" = "SG" * "1.0 g mL}}^{- 1}$

${\rho}_{\text{solution" = "SG" quad "g mL}}^{- 1}$

So, you know that $\text{1 mL}$ of this solution has a mass of $\left(\text{SG}\right)$ $\text{g}$, so pick a $\text{100-g}$ 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 $x$ $\text{g}$ of the solute, so use the molar mass of the solute, let's say ${M}_{M}$ ${\text{g mol}}^{- 1}$, 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 $\text{1 L" = 10^3 quad "mL}$ of the solution.

Since $\left(\frac{100}{\text{SG}}\right)$ $\text{mL}$ of this solution contain $\left(\frac{x}{M} _ M\right)$ moles of solute, you can say that ${10}^{3}$ $\text{mL}$ 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 $\text{SG}$--with the density of water equal to ${\text{1.0 g mL}}^{- 1}$--a percent concentration by mass equal to x%#, and a solute that has a molar mass of ${M}_{M}$ ${\text{g mol}}^{- 1}$ is equal to

${\text{molarity" = ((10 * x * "SG")/M_M) quad "moles L}}^{- 1}$

This solution contains $\left(\frac{10 \cdot x \cdot \text{SG}}{M} _ M\right)$ moles of solute for every $\text{1 L" = 10^3 quad "mL}$ of the solution.