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Chapter 9: Q. 9.18 (page 294)

An intravenous solution to replace potassium loss contains $40 m E q / L$ each of $K^{+}$ and $C l^{-}$ . How many moles each of $K^{+}$ and $C l^{-}$ are in $1.5 L$ of the solution?

Short Answer

Expert verified

In $1.5 L$ of the solution, moles of $K^{+}$ and $C l^{-}$ are $0.06$ and $0.06$ respectively.

Step by step solution

01

Given Information.

Potassium loss contains $40 m E q / L$

Solution is $1.5 L$

To find the moles present in the solution.

02

Find the equivalent.

$1 L$ of intravenous solution contains $40 m E q / L$

Therefore, convert $m E q$ to $E q$ as follows:

$1 Eq = 1000 mEq = 40 m E q 脳 \frac{1 Eq}{1000 mEq} = 0.04 E q$

03

Find the moles in K+

Charge of Potassium is $+ 1$ . Therefore,

$1 E q o f K^{+} = 1 m o l e o f K^{+}$

$= 1.5 L 脳 \frac{0.04 Eq of K^{+}}{1 L} 脳 \frac{1 mol of K^{+}}{1 Eq of K^{+}} = 0.06 m o l e o f K^{+}$

Hence, mole of $K^{+}$ ions is $0.06 m o l e s$

04

Find the moles in Cl-

Charge of Chlorine is $- 1$ . Therefore,

$1 E q o f C l^{-} = 1 m o l e o f C l^{-}$

Then,

$= 1.5 L 脳 \frac{0.04 Eq of {Cl}^{鈭�}}{1 L} 脳 \frac{1 mol of {Cl}^{鈭�}}{1 Eq of {Cl}^{鈭�}} = 0.06 m o l e s o f C l^{-}$

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Most popular questions from this chapter

Calculate the mass/volume percent $(m / v)$ for the solute in each of the following:

a. $75 g o f N a_{2} S O_{4}$ in $250 m L$ of $N a_{2} S O_{4}$ solution.

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d. urea solution and $N a C l$ solution.

Calculate the number of equivalents in each of the following:

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c. $1 {mole ofCa}^{2 +}$

d. ${3molesofCO}_{3}^{2 -}$

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How many liters of a $4.00 M NaCl$ solution will provide $25.0 g$ of $NaCl ?$

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