To prepare 250. ml of a 0.125 m solution of solid KCl are needed: 1.125 grams of solid KCl
To solve this problem the formulas and the procedures that we have to use are:
- m = n(sto)/v(soution) L
- MW= ∑ AWT
- n = m / MW
Where:
- n = moles
- MW = molecular weight
- AWT = atomic weight
- m = molar concentration
- v = volume
Information about the problem:
- m = 0.125 mol/L
- v(solution) = 250 ml
- AWT (K)= 19 g/mol
- AWT (Cl) = 17 g/mol
- m(kcl) =?
By converting the volume units from (ml) to (L) we have:
v(solution) = 250 ml * 1 L/1000 ml
v(solution) = 0,250 L
Applying the molar concentration formula and clearing the mole (n) we get:
m= n(sto)/v(soution) L
n(KCl) = m * v(soution) L
n(KCl) = 0.125 mol/L * 0.250 L
n(KCl) = 0.03125 mol
Applying the molecular weight formula we get:
MW= ∑ AWT
MW(KCl)= 19 g/mol + 17 g/mol
MW(KCl)= 36 g/mol
Applying the mole (n) formula and clearing the mass (m) we get:
n = m / MW
m = n * MW
m(KCl) = n(KCl) * MW(KCl)
m(KCl) = 0.03125 mol * 36 g/mol
m(KCl) = 1.125 g
What is a solution?
In chemistry a solution is known as a homogeneous mixture of two or more components called:
Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161
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