Question

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

Short answer

The solution is $204mlLiCl$of$1.75MLiCl$solution contains$15.2g$of$LiCl$.

Step-by-step solution

Given information

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

Solution for given data:

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

Equalities /Conversion factors:

$1moleLiCl$role="math" localid="1652512944024" $=42.5\mathrm{g}\mathrm{LiCl}$

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

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

role="math" localid="1652513781092" $\begin{array}{cll}1\mathrm{LLiCl}& =1.75molesLiCl& \\ \frac{1\mathrm{LLiCl}}{1.75\mathrm{moleLiCl}}& \text{and}\frac{1.75\mathrm{moleLiCl}}{1\mathrm{LLiCl}}& \end{array}$

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

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

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

Final calculation :

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

$=204mlLiCl.$