Question #8632c

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Question #8632c

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Answer 1 · 2017-05-02T15:58:48

$Molarity = 0.10 \cdot m o l \cdot L^{- 1}$ $"Molarity"=0.10*mol*L^-1$

Explanation:

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

And thus, here:

$Molarity = \frac{\frac{17.1 \cdot g}{342.30 \cdot g \cdot m o l^{- 1}}}{0.50 \cdot L} = 0.10 \cdot m o l \cdot L^{- 1}$ $"Molarity"=((17.1*g)/(342.30*g*mol^-1))/(0.50*L)=0.10*mol*L^-1$ 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 $m o l \cdot L^{- 1}$ $mol*L^-1$ ............i.e. $\frac{1}{m o l^{- 1}} = \frac{1}{\frac{1}{m o l}} = m o l$ $1/(mol^-1)=1/(1/(mol))=mol$ as required...........

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Question #8632c

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