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22 June, 12:34

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

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  1. 22 June, 12:55
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    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
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