Answer:
a) Molarity KCl = 0.755 M
b) molality HNO = 5.09 m
Explanation:
- Formality (F) = moles sto / L sln
- Molarity (M) = # dissolved specie / L sln
- molality (m) = moles sto / Kg ste
- %p/p = ( g sto / g ste ) * 100
a) KCl ↔ K+ + Cl-
moles KCL:
⇒ 20 g KCl * ( mol / 74.6 g ) = 0.268 mol KCl
⇒ F = 0.268 mol KCl / 0.355 L = 0.755 F
⇒ M [ K+ ] = 1 * ( 0.755) = 0.755 M
b) 24% HNO:
calculation base: 1 g solution:
⇒24 = ( g sto / g sln) * 100
⇒ 0.24 = g sto / 1
⇒ g sto = 0.24g
⇒g sln = 1 - 0.24 = 076g sln
⇒Kg ste = 0.76 g * ( Kg / 1000g ) = 7.6 E-4 Kg ste
moles sto (HNO):
⇒ 0.24g * ( mol / 62.03g) = 3.869 E-3 moles HNO
⇒ m = 3.869 E-3 moles HNO / 7.6 E-4 Kg sln
⇒ m = 5.09 mol/Kg
