What is the molarity of a 50% rubidium hydroxide solution #"(w/w)"# whose density is #1.74*g*mL^-1#?

1 Answer
Apr 20, 2017

Answer:

Now we are given that #"Mass of solute"/"Mass of solution"xx100%=50%#.

Explanation:

We are further given #rho_"RbOH solution"=1.74*g*mL^-1#.

We know that the #"molar mass"# of #RbOH=102.48*g*mol^-1#. (And, in fact, as I recall, you can only buy this as an aqueous solution.)

And we can work out the molar concentration on the basis of these data, given a #1*mL# volume of #RbOH(aq)#

#"Concentration"# #=# #"Moles of RbOH"/"Volume of solution (L)"#

#=((1*cancel(mL)xx1.74*cancelg*cancel(mL^-1)xx50%)/(102.48*cancelg*mol^-1))/(1xx10^-3*L)#

#=8.50*mol*L^-1#

This is dimensionally consistent; the calculation gave units of #1/(mol^-1)xxL^-1=1/(1/(mol))xxL^-1=mol*L^-1# as required.