Question

A `"98.8% w/w"` solution of sulfuric acid has `rho=1.84*g*mL^-1`. What is the molar concentration of the acid?

Answer 1

I take it you want to make a `1*L` volume of a `3.0*mol*L^-1` solution.....

Explanation:

And, as I have said before, there is a very important practical issue to be considered when we dilute a strong acid. Because addition of a strong acid to water results in an EXOTHERM, i.e. an evolution of heat as the acid is hydrated, we ALWAYS ADD ACID to WATER, and NEVER the reverse....

i.e. `"when you spit in acid it spits back, so don't spit in the acid"`

And now finally to business...

We use the relationship...`"concentration"="moles of solute"/"volume of solution"`

And thus `"moles of solute"="volume of solution"xx"concentration"`

Here this gives ........`"moles of solute"=1*Lxx3.0*mol*L^-1=3.0*mol`

And so from the conc. acid...we need...`3.0*molxx98.08*g*mol^-1=294.2*g` WITH RESPECT TO SOLUTE....

And this is a volume of.....`(294.2*g)/(98%xx1.84*g*mL^-1)=163.2*mL`, which is then diluted to a `1*L` volume with water.

When you do the dilution you must be togged up in the safety gear: safety specs or optical specs, and a lab coat to protect your clothes. This is an absolute requirement.....

For your information quotient..... `1*cm^3=1xx(10^-2*m)^3=1xx10^-6*m^3=1/1000xx1/1000xx1*m^3=1/1000xx1*L=10^-3L=1*mL................` as required. The point I try to make is that `cm^3` and `mL` are the SAME unit of volume.