91Ó°ÊÓ

The molarity of a solution is just the molar mass of solute inside one liter of solution. Molarity is mathematically expressed as

$\text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}$

Given:

$2.00$moles of a glucose solution in $4.00L$of a glucose solution

Substitute the values in formula,

$$\begin{gathered} \text{Molarity}=\frac{\text{moles of glucose}}{\text{liters of solution}} \\ =\frac{2.00\text{moles}}{4.00\mathrm{L}} \\ =0.500M \end{gathered}$$

Given:

$4.00gofKOH$in $2.00LofaKOH$solution.

Find the moles for localid="1652697221540" $4.00gofKOH,$

$1\text{mole of}\mathrm{KOH}=56.1\mathrm{g}$

And then,

localid="1653894874376" $\frac{1\text{mole}\mathrm{KOH}}{56.1g\mathrm{KOH}}\text{and}\frac{56.1g\mathrm{KOH}}{1\text{mole}\mathrm{KOH}}$

Hence, Convert it to find the moles for grams,

localid="1653894885035" $$\begin{gathered} \text{Moles of}\mathrm{KOH}=4.00\mathrm{g}KOH×\frac{1\text{moles}\mathrm{KOH}}{56.1g\mathrm{KOH}} \\ =0.0710mole \end{gathered}$$

Substitute the values in molarity formula,

localid="1653894903560" $$\begin{gathered} \text{Molarity}=\frac{\text{moles of glucose}}{\text{liters of solution}} \\ =\frac{0.071\text{moles}}{2.00\mathrm{L}} \\ =0\text{.0357M} \end{gathered}$$

Given:

$5.85gofNaCl$in $400mLofaNaCl$solution.

Find the moles for $5.85gofNaCl,$

localid="1653894914657" $1\text{mole of}\mathrm{NaCl}=58.4\mathrm{g}$

And then,

localid="1652697571694" $\frac{1\text{mole}\mathrm{NaCl}}{58.4g\mathrm{NaCl}}\text{and}\frac{58.4g\mathrm{NaCl}}{1\text{mole}\mathrm{NaCl}}$

Hence, Convert it to find the moles for grams,

localid="1653894927542" $$\begin{gathered} \text{Moles of}\mathrm{NaCl}=5.85g\mathrm{NaCl}×\frac{1\text{moles}\mathrm{NaCl}}{58.4g\mathrm{NaCl}} \\ =0.100M \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{Molarity}=\frac{\text{moles of}\mathrm{NaCl}}{\text{liters of solution}} \\ =\frac{0.100\text{moles}}{400\mathrm{mL}×\frac{1\mathrm{L}}{1000\mathrm{mL}}} \\ =0.250M \end{gathered}$$