Specific gravity of HCl is 1.08. what is its Molarity???
Here's what I got.
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
A hydrochloric acid solution that has a density of
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")#