What is the molarity of a 50% rubidium hydroxide solution $"(w/w)"$ whose density is $1.74gmL^-1$ ?

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Answer 1 · 2017-04-20T13:57:24

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

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.

What is the molarity of a 50% rubidium hydroxide solution "(w/w)""(w/w)" whose density is 1.74*g*mL^-11.74*g*mL^-1?