# Specific gravity of HCl is 1.08. what is its Molarity???

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Before doing anything else, make sure that you have a clear understanding of what *specific gravity* means.

**Specific gravity** usually refers to the ratio between the *density* of a given substance and the density of a reference substance at the same temperature. More often than not, that reference substance is **water**.

#color(blue)("SG" = rho_"substance"/rho_"water")#

Since no mention of temperature was made, you can assume the density of water to be equal to

In this context, hydrochloric acid's specific gravity tells you what the *density* of a specific hydrochloric acid solution is **compared** with that of water.

More specifically, the density of your hydrochloric acid solution will be equal to

#"SG" = rho_"HCl"/rho_"water" implies rho_"HCl" = "SG" xx rho_"water"#

#rho_"HCl" = 1.08 xx "1.00 g/mL" = "1.08 g/mL"#

So, you know what the density of the solution is. To find its molarity, you need to pick a sample of this solution and figure out how many **moles** of hydrochloric acid it contains.

Now, you're going to need the percent concentration by mass of hydrochloric acid in this solution, which you can look up here

https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration

A hydrochloric acid solution that has a density of **for every**

To make the calculations easier, pick a

#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.08 g"/(1color(red)(cancel(color(black)("mL")))) = "1080 g"#

Next, use its percent concentration by mass to determine how much *hydrochloric acid* it contains

#1080color(red)(cancel(color(black)("g solution"))) * "17 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "183.6 g"#

Next, use hydrochloric acid's molar mass to determine how many *moles* of hydrochloric acid you have in this many grams

#183.6color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "5.0357 moles HCl"#

Since molarity is defined as moles of solute, which in your case is hydrochloric acid, divided by **liters** of solution, and you picked an initial sample of

#color(blue)(c = n/V)#

#c = "5.0357 moles"/"1.00 L" = "5.0357 M"#

Rounded to three sig figs, the number of sig figs you have for the solution's specific gravity, the answer will be

#c = color(green)("5.04 M")#