Chapter U4, Problem 10STP

Interpretation Introduction

Interpretation:

The molarity of $\text{45}{\text{.0 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}$ dissolved in $\text{250 mL}$ of water needs to be determined.

Concept introduction:

The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

$\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 10STP

The molarity of $\text{45}{\text{.0 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}$ dissolved in $\text{250 mL}$ of water is $\text{0}\text{.53 M}$.

Explanation of Solution

The formula to calculate number of moles is:

$\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$

Now, equation (1) an be rewritten as:

$\text{Molarity = }\frac{\frac{\text{mass of the solute}}{\text{Molar mass of the solute}}}{\text{Volume of solution (L)}}$ - (2)

The molar mass of sugar, ${\text{C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}$ is $\text{342}\text{.3 g/mol}$.

Converting the given volume of sample from mL to L as:

$250\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}=\text{ 0}\text{.25 L}$

Substituting the given values in equation (2) to determine the molarity of the solution as:

$\begin{array}{l}\text{Molarity = }\frac{\frac{\text{45 g}}{\text{342}\text{.3 g/mol}}}{\text{0}\text{.25 L}}\\ \text{Molarity = }\frac{\text{0}\text{.1315 mol}}{\text{0}\text{.25 L}}\\ \text{Molarity = 0}\text{.526 mol/L }\simeq \text{ 0}\text{.53 M}\end{array}$

Since, the molarity of $\text{45}{\text{.0 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}$ dissolved in $\text{250 mL}$ of water is $\text{0}\text{.53 M}$ so, the correct option is (a).