Q. 9.101 - An Introduction to General, Organic, and Biological Chemistry

Question

How many milliliters of a $1.75 \mathrm{M} LiCl$ solution contain $15.2 \mathrm{g}$ of $LiCl$? 

Step-by-Step Solution

Answer

The solution is $204 ml LiCl$of $1.75 M LiCl$solution contains $15.2 g$of $LiCl$.

Step 1: Given information

To find how many milliliters of $1.75 M LiCl$ solution contains $15.2 g$of $LiCl$:

Step 2: Solution for given data:

Gram $LiCl$Molar mass moles $LiCl$ Molarity $LiCl$ metric system $ml LiCl$.

Step 3: Equalities /Conversion factors:

$1mole LiCl$ $= 42.5 \mathrm{g} \mathrm{LiCl}$

$\frac{1\mathrm{moleLiCl}}{42.5 \mathrm{gLiCl}}$ and $\frac{42.5 \mathrm{gLiCl}}{1 \mathrm{moleLiCl}}$

The molarity of any solution can be written as two conversion factors 

$\begin{array}{cll}1\mathrm{LLiCl}& =1.75 moles LiCl& \\ \frac{1\mathrm{LLiCl}}{1.75 \mathrm{moleLiCl}}& \text{ and } \frac{1.75\mathrm{moleLiCl}}{1\mathrm{LLiCl}}& \end{array}$

The liter of solution can be converted into milliliter of solution using the metric system

$1000 ml \mathrm{LiCl}=1 L LiCl$

$\frac{1000\mathrm{mlLiCl}}{1\mathrm{LLiCl}}$and $\frac{1\mathrm{LLiCl}}{1000\mathrm{mlLiCl}}$

Step 4: Final calculation :

$\mathrm{ml}$ of $\mathrm{LiCl}=15.2 \mathrm{g} \mathrm{LiCl}\times \frac{1\text{ mole LiCl }}{42.5 \mathrm{g} \mathrm{LiCl}}\times \frac{1 \mathrm{LiCl}}{1.75\mathrm{mole} L\mathrm{iCl}}\times \frac{1000\mathrm{ml} \mathrm{LiCl}}{1 \mathrm{L} \mathrm{LiCl}}$

$=204 ml LiCl.$