Find the molarity and molality of 37.0 wt% hcl. assume the density of the acid is 1.19g/ml

a) M HCl = 32.162 mol/L

b) m HCl = 15.873 mol/Kg

Explanation:

wt% = (mass HCl / mass sln)×100δ HCl = mass HCl / Volume slnmolarity = 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

I’m pretty sure it’s false i’m not sure

Sorry i think the answer is b