Answer:
a) M HCl = 32.162 mol/L
b) m HCl = 15.873 mol/Kg
Explanation:
- wt% = (mass HCl / mass sln)×100
- δ HCl = mass HCl / Volume sln
- molarity = mol HCl / V sln (L)
- molality = mol HCl / Kg solvent
∴ MW HCl = 36.46 g/mol
assuming mass sln = 1 g
∴ mass sln = 1 g = g solvent + g solute
∴ solute: HCl
⇒ 0.37 = g HCl / g sln
⇒ g HCl = 0.37 g
⇒ g ste = 1 - 0.37 = 0.63 g
⇒ Kg ste = (0.63 g)×(Kg/1000 g) = 6.3 E-4 Kg
⇒ mol HCl = (0.37 g HCl)×(mol/36.46 g HCl) = 0.010 mol HCl
∴ V sln = (0.37 g)/(1.19 g/mL) = 0.3109 mL
⇒ V sln = (0.31 mL)×( L/1000 mL) = 3.109 E-4 L
a) molarity (M):
⇒ M HCl = 0.010 mol HCl / 3.109 E-4 L
⇒ M HCl = 32.162 mol/L
b) molality (m):
⇒ m HCl = ( 0.010 mol HCl) / (6.3 E-4 Kg)
⇒ m HCl = 15.873 mol/Kg
