Question

Calculate the molarity of each of the following:

a.$0.500$mole of glucose in a$0.200L$of a glucose solution.

b. $73.0gofHCl$in a $2.00LofHCl$solution.

c. $30.0gofNaOH$in$350mLofNaOH$solution.

Short answer

(part a) Molarity of $0.500$mole of glucose in a $0.200L$of glucose solution is $2.5M.$

(part b) Molarity of $73.0gofHCl$in $2.00LofHCl$solution$1.0M.$

(part c) Molarity of$30.0gofNaOH$in$350mLofNaOH$solution is$2.14M.$

Step-by-step solution

Introduction (part a).

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

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

Given Information (part a). 

Given:

$0.500$mole of glucose in a $0.200L$of a glucose solution.

Explanation (part a). 

Substitute the values in formula,

$$\begin{gathered} \mathrm{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}} \\ =\frac{0.500\mathrm{mol}\text{glucose}}{0.200\mathrm{L}\text{water}} \\ =2.5mol/L \\ =2.5M \end{gathered}$$

Given Information (part b). 

Given:

$73.0gofHCl$in$2.00LofHCl$solution.

Explanation (part b). 

Find the moles for $73.0gofHCl.$

$$\begin{gathered} \text{number of moles of}\mathrm{HCl}=\text{mass of}\mathrm{HCl}\times \frac{1\mathrm{mol}\mathrm{HCl}}{\text{molar mass of}\mathrm{HCl}} \\ =73.0\mathrm{g}\times \frac{1\mathrm{mol}\mathrm{HCl}}{36.46\mathrm{g}} \\ =2.00molHCl \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{Molarity}\left(\mathrm{M}\right)\text{of}\mathrm{HCl}=\frac{2.00\mathrm{mol}\mathrm{HCl}}{2.00\mathrm{L}\text{water}} \\ =1.0\mathrm{mol}/\mathrm{L} \\ =1.0\mathrm{M}\mathrm{HCl} \end{gathered}$$

Given Information (part c). 

Given:

$30.0gofNaOH$in$350mLofNaOH$solution.

Explanation (part c). 

Find the moles for $30.0gofNaOH,$

$$\begin{gathered} \text{number of moles ofNaOH}=\text{mass of}\mathrm{NaOH}\times \frac{1\mathrm{mol}\mathrm{NaOH}}{\text{molar mass of}\mathrm{NaOH}} \\ =30.0\mathrm{g}\times \frac{1\mathrm{mol}\mathrm{HCl}}{39.997\mathrm{g}} \\ =0.750mol\mathrm{NaOH} \end{gathered}$$

Covert milliliter to liter,

$$\begin{gathered} 1000mL=1L \\ 350mL\times \frac{1.00\mathrm{L}}{1000\mathrm{mL}}=0.350\mathrm{L} \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{Molarity}\left(\mathrm{M}\right)\text{ofNaO}H=\frac{2.00\mathrm{mol}\mathrm{NaO}H}{2.00\mathrm{L}\text{water}} \\ =2.14\mathrm{mol}/\mathrm{L} \\ =2.14\mathrm{M}\mathrm{HCl} \end{gathered}$$