A solution of #H_2SO_4(aq)# with a molal concentration of 3.58 m has a density of 1.200 g/mL. What is the molar concentration of this solution?

Question

Ellie Andrews · Accepted Answer

#"3.18 M"#

Knowing the molal concentration, here is what you know... by assuming #"1 kg"# of aqueous solvent, we have:

#"3.58 mols H"_2"SO"_4#

#"1 kg water"#

By using the density of the solution, along with the mass of the solution, we can get the volume of the solution. Then, by definition, the molarity is #"mols solute"/"L solution"#.

#m_(soln) = 3.58 cancel("mols H"_2"SO"_4) xx ("98.079 g H"_2"SO"_4)/cancel("1 mol H"_2"SO"_4) + cancel"1 kg water" xx "1000 g"/cancel"1 kg"#

#= 1.35_(112282) xx 10^(3) "g soln"#

As a result, its volume is:

#1.35_(112282) xx 10^(3) cancel"g soln" xx cancel"1 mL soln"/(1.200 cancel"g soln") xx "1 L soln"/(1000 cancel"mL soln")#

#= 1.12_(5935683)# #"L soln"#

Its molar concentration is thus:

#("3.58 mols H"_2"SO"_4)/(1.12_(5935683) "L soln")#

#=# #color(blue)("3.18 M")#

Peyton Adams · Answer

undefined To find the molar concentration of the solution, we need to use the formula:

$$ \text{Molar concentration} = \text{Molal concentration} \times \text{Density} $$

Given that the molal concentration ($m$) is $3.58$ m and the density is $1.200$ g/mL, we can plug these values into the formula:

$$ \text{Molar concentration} = 3.58 \, \text{m} \times 1.200 \, \text{g/mL} $$

Since the density is in grams per milliliter, we need to convert it to grams per liter ($g/L$) by multiplying by $1000$:

$$ \text{Molar concentration} = 3.58 \, \text{m} \times (1.200 \, \text{g/mL} \times 1000 \, \text{mL/L}) $$

Now, we can calculate the molar concentration.