# Question #8632c

May 2, 2017

$\text{Molarity} = 0.10 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

By definition, $\text{Molarity"="Moles of solute"/"Volume of solution (L)}$..

And thus, here:

$\text{Molarity} = \frac{\frac{17.1 \cdot g}{342.30 \cdot g \cdot m o {l}^{-} 1}}{0.50 \cdot L} = 0.10 \cdot m o l \cdot {L}^{-} 1$ with respect to sucrose.

So all I had to do was solve this quotient, and preserve the units of the quotient to give an answer in $m o l \cdot {L}^{-} 1$............i.e. $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$ as required...........