Q. 9.17 - An Introduction to General, Organic, and Biological Chemistry

Question

An intravenous saline solution contains $154 mEq/L$ each of $N{a}^{+}$and $C{l}^{-}$. How many moles each of $N{a}^{+}$and $C{l}^{-}$ are in $1.00 L$ of the saline solution?

Step-by-Step Solution

Answer

Moles of $N{a}^{+}$and $C{l}^{-}$ in  $1.00 L$ of saline water are $0.154, 0154$ respectively.

Step 1: Given Information.

Saline solution contains $154 mEq/L$

To find mole in $1.00L$ of saline water.

Step 2: Find the Equivalent.

The sodium content of salt solution is percent sodium (salt), which is similar to the sodium content of blood and tears. Saline solution is generally known as normal saline, but is also known as physiological or isotonic saline.

$1 L$ of the provided saline solution includes $154 mEq$ of $N{a}^{+}$ and $C{l}^{-}$ions, respectively. To start, transform $mEq$ to $Eq$as follows:

$$\begin{gathered} 1\mathrm{Eq}=1000\mathrm{mEq} \\ =154 mEq\times \frac{1 Eq}{1000 mEq} \\ =0.154 Eq \end{gathered}$$

In $1 L$ of saline water, $1.154 Eq$ of $N{a}^{+}$ and $C{l}^{-}$ ions are present.

Step 3: Find the moles of N a +

Charge of Sodium is $+1$

Therefore,

$1 Eq of N{a}^{+}=1 mole of N{a}^{+}$

$$\begin{gathered} =1 \mathrm{L}\times \frac{0.154\mathrm{Eq}\text{ of }{\mathrm{Na}}^{+}}{1 \mathrm{L}}\times \frac{1 \mathrm{mol}\text{ of }{\mathrm{Na}}^{+}}{1\mathrm{Eq}\text{ of }{\mathrm{Na}}^{+}} \\ =0.154 mole of N{a}^{+} \end{gathered}$$

Therefore, the mole of $N{a}^{+}=0.154$

Step 4: Find the mole of C l -

Charge of Chlorine is $-1$. so,

$1 Eq of C{l}^{-}=1 mole of C{l}^{-}$

$$\begin{gathered} =1 \mathrm{L}\times \frac{0.154\mathrm{Eq}\text{ of }{\mathrm{Cl}}^{-}}{1 \mathrm{L}}\times \frac{1 \mathrm{mol}\text{ of }{\mathrm{Cl}}^{-}}{1\mathrm{Eq}\text{ of }{\mathrm{Cl}}^{-}} \\ =0.154 mol of C{l}^{-} \end{gathered}$$

Hence, the moles present in $C{l}^{-}$ ion is $0.154$