Question

A solution to replace potassium loss contains 43 mEq/L each of K+and Cl-

1- how many moles of K+ are in 1.9 L of the solution?

_________\_= mole K+

2- how many moles of Cl- are in 1.9 L of the solution

__________=molCl-

Express your answer using two sig figs

Answer 1

equivalent weight of K+ and Cl- is equal to their molar mass
so,
number of mole = number of equivalent

number of milliequivalent = (concentration in milliequivalent)*(volume in L)
1)
number of milliequivalent of K+ = 43*1.9
= 81.7 milliequivalent

number of equivalent of K+ = (81.7*10^-3) Eq
= (8.17*10^-2) mole
= (8.2*10^-2) mole
2)
number of milliequivalent of Cl- = 43*1.9
= 81.7 milliequivalent

number of equivalent of Cl- = (81.7*10^-3) Eq
= (8.17*10^-2) mole
= (8.2*10^-2) mole