Question

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

Short answer

Moles of $N{a}^{+}$and$C{l}^{-}$inrole="math" localid="1652535692509" $1.00L$of saline water are$0.154,0154$respectively.

Step-by-step solution

Given Information.

Saline solution contains $154mEq/L$

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

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.

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

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

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

Find the moles of Na+

Charge of Sodium is $+1$

Therefore,

$1EqofN{a}^{+}=1moleofN{a}^{+}$

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

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

Find the mole of Cl-

Charge of Chlorine is $-1$. so,

$1EqofC{l}^{-}=1moleofC{l}^{-}$

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

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