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

Apr 20, 2017

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

#### Explanation:

We are further given ${\rho}_{\text{RbOH solution}} = 1.74 \cdot g \cdot m {L}^{-} 1$.

We know that the $\text{molar mass}$ of $R b O H = 102.48 \cdot g \cdot m o {l}^{-} 1$. (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 $1 \cdot m L$ volume of $R b O H \left(a q\right)$

$\text{Concentration}$ $=$ $\text{Moles of RbOH"/"Volume of solution (L)}$

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

$= 8.50 \cdot m o l \cdot {L}^{-} 1$

This is dimensionally consistent; the calculation gave units of $\frac{1}{m o {l}^{-} 1} \times {L}^{-} 1 = \frac{1}{\frac{1}{m o l}} \times {L}^{-} 1 = m o l \cdot {L}^{-} 1$ as required.