Q150CP - Chemistry: Molecular Nature Of Matter And Change

Question

The total concentration of dissolved particles in blood is $\begin{array}{l}\mathbf{\text{0.30 M}}\end{array}$. An intravenous (IV) solution must be isotonic with blood, which means it must have the same concentration.

(a) To relieve dehydration, a patient is given $\begin{array}{l}\mathbf{\text{100 }}\raisebox{1ex}{$\mathbf{\text{mL}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{h}}$}\right.\end{array}$ of IV glucose $\mathbf{\left(}{\mathbf{C}}_6{\mathbf{H}}_{12}{\mathbf{O}}_6\mathbf{\right)}$ for $\mathbf{\text{2.5 h}}$. What mass $\left(\text{g}\right)$ of glucose did she receive?

(b) If isotonic saline $\left(\text{NaCl}\right)$ is used, what is the molarity of the solution?

(c) If the patient is given $\mathbf{\text{150 }}\raisebox{1ex}{$\mathbf{\text{mL}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{h}$}\right.$ of IV saline for $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{h}$, how many grams of $\mathbf{NaCl}$ did she receive?

Step-by-Step Solution

Answer

(a) She received$13.51\text{ g}$ mass of glucose.

(b) Molarity of the saline solution is $0.15\text{ M}$.

(c) She received $1.97\text{ g}$ of $\mathrm{NaCl}$.

A concept:

The number of moles of solute present in a specific number of liters of the solution, or moles per liter of a solution, is known as molar concentration or molarity.

(a) Mass of glucose:

Given that the concentration of dissolved particles in blood is $0.30\text{ M}$ and the intravenous (IV) solution must be isotonic with blood. $100\text{ mL}$ of IV glucose are administered to the patient each hour.

IV glucose dosage administered over $2.5\text{ hours}$ is, 

$\frac{100\text{ mL}}{1\text{ hour}}\times 2.5\text{ hours}=250\text{ mL}$ 

 

The glucose solution has a concentration of $0.30\text{ M}$.

 

Mass of glucose can be calculated by using following formula.

$\text{molarity}=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}glucose}}{\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}glucose}}\times \frac{1}{V\left(\text{L}\right)}$ 

$\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}glucose}& =& \text{molarity}\times \text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}gluocse}\times V\left(\text{L}\right)\\ & =& 0.30\text{ M}\times 180.156\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times \frac{250\text{ mL}}{1000\text{ L}}\\ & =& 13.51\text{ g}\end{array}$

 

Thus, the mass of glucose received by the patient in $\text{2.5 h}$ is $\text{13.51 g}$.

(b) Molarity of the solution:

Molarity is defined as the moles of solute dissolved in per liter of solution.

$\text{NaCl}$ is a strong electrolyte. When dissolves, it gives $\text{2 mol}$ of ions. To make isotonic solution of $\text{NaCl}$ with $\text{0.30 M}$ concentration of blood, the molarity of saline solution should be half of its molarity.

$\begin{array}{rcl}\text{molarity\hspace{0.17em}of NaCl}& =& \frac{\text{concentration\hspace{0.17em}of\hspace{0.17em}blood}}{2}\\ & =& \frac{0.30{\displaystyle \text{ M}}}{2}\\ & =& 0.15\text{ M}\end{array}$

Therefore, $0.15\text{ M}$ of $\text{NaCl}$ will produce $0.30\text{ moles}$ of ions in the solution which will be isotonic to blood.

(c) Mass of NaCl :

Calculated concentration of dissolved ions of $\text{NaCl}$ in blood is $\text{0.15 M}$.

Volume of IV saline given to the patient per hour is $\text{150 mL}$.

 

Volume of IV saline given in $\text{1.5 hours}$ is,

$\frac{150\text{ mL}}{1\text{ hour}}\times 1.5\text{ hours}=225\text{ mL}$ 

Concentration of the saline solution is $0.15\text{ M}$.

 

Mass of $\text{NaCl}$ can be calculated by using following formula.

$\text{molarity}=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}NaCl}}{\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}NaCl}}\times \frac{1}{V\left(\text{L}\right)}$ 

$\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}NaCl}& =& \text{molarity}\times \text{molar\hspace{0.17em}}NaCl\times V\left(\text{L}\right)\\ & =& 0.15\text{ M}\times 58.44\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times \frac{\text{225 mL}}{\text{1000 L}}\\ & =& 1.97\text{ g}\end{array}$

 

Hence, the mass of NaCl received by the patient in $1.5\text{ h}$ is $1.97\text{ g}$.