Question #9e481

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Question #9e481

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Answer 1 · 2014-12-09T18:38:21

The answer is $4.0 g$ $4.0g$ .

We know that molarity is the concentration of a solution expressed by the number of moles of solute per liter of solution. This means that

$C = \frac{n_{s o l u t e}}{V_{s o l u t i o n}} \to n_{H C l} = C \cdot V_{s o l u t i o n}$ $C = n_(solute)/(V_(solution)) -> n_(HCl) = C * V_(solution)$

So, $n_{H C l} = 1.5 \frac{m o l e s}{L} \cdot 75 \cdot 10^{- 3} L = 0.11 m o l e s$ $n_(HCl) = 1.5 (mol es)/(L) * 75 * 10^(-3) L = 0.11mol es$

Knowing that $H C l$ $HCl$ 's molar mass is equal to $36.5 \frac{g}{m o l}$ $36.5 g/(mol)$ , we can determine its mass by

$m_{H C l} = n_{H C l} \cdot 36.5 \frac{g}{m o l} = 0.11 m o l e s \cdot 36.5 \frac{g}{m o l e} = 4.0 g$ $m_(HCl) = n_(HCl) * 36.5 g/(mol) = 0.11 mol es * 36.5 g/(mol e) = 4.0 g$

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Question #9e481

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