. How many grams of potassium chloride, KCl, are needed to prepare 0.630 L of a 2.50 M solution of potassium chloride? (4.0 points) (Points :4)
1.575 M
117 g
2.54 m
23.19 g
solve by
molarity =no. of moles divide by volume in litres.......
so let the mass be in x grms...
then :
2.50*0.630=x grms
so,
x grms of KCl =1.575 grams is the answer
To determine the number of grams of potassium chloride (KCl) needed to prepare the given solution, we can use the formula:
Moles of solute = Molarity x Volume (in liters)
First, let's calculate the moles of potassium chloride needed:
Moles of solute = 2.50 M x 0.630 L = 1.575 moles
To convert moles to grams, we need to know the molar mass of potassium chloride (KCl). The molar mass of K is 39.10 g/mol, and the molar mass of Cl is 35.45 g/mol.
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
Now, we can calculate the grams of potassium chloride needed:
Grams of KCl = Moles of solute x Molar mass of KCl
= 1.575 moles x 74.55 g/mol
= 117.26 g (approximately)
Therefore, the correct answer is 117 g.