How many grams of LiOH would be needed to make a 500.0 mL of 0.340 M LiOH?

Question:

How many grams of LiOH would be needed to make a 500.0 mL of 0.340 M LiOH?

Preparing Reagents:

In Chemistry experiments, we prepare our own solutions. In order to acquire the desired solution with a certain volume and concentration, we must know the number of moles of a substance, and then multiply it with the molar mass to determine how much of the reagent must be weighed.

Answer and Explanation:

Determine the mass of {eq}\displaystyle LiOH {/eq} to acquire the given solution by first finding the number of moles there is in the solution, by multiplying the concentration to the volume, {eq}\displaystyle n = MV {/eq}, and then multiplying it to the molar mass of {eq}\displaystyle LiOH {/eq}, which is {eq}\displaystyle 23.95\ \rm{g/mol} {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle m_{LiOH} = n\times 23.95\ \rm{g/mol} = 0.5000\ L \times 0.340\ M\ times 23.95\ \rm{g/mol} =4.07\ g\ LiOH \end{align} {/eq}


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