Answer to Question #314818 in General Chemistry for Loeeyyy

Question

1. What is the molar concentration of Aki’s disinfectant if she dissolved 5Tbsp. of

NaClO (sodium hypochlorite) in 3.8 liters of H2O (water)?

2. Compute for the molality of her disinfectant if she dissolved 5Tbsp. of NaClO

(sodium hypochlorite) in 3.8 liters of H2O (water).

3. Calculate the percent by mass of sodium hypochorite in her disinfectant solution. (

5Tbsp. of NaClO (sodium hypochlorite) and 3.8 liters of H2O (water). In units of grams

for both of the solute and solvent.

4. Calculate the mole fraction of sodium hypochorite and water in Aki’s solution. (

5Tbsp. of NaClO (sodium hypochlorite) and 3.8 liters of H2O (water).

5. Calculate the percent by volume of the disinfectant Aki made. Units in mL.

6. What is the concentration of her solution in parts per million? ( 5Tbsp. of NaClO

(sodium hypochlorite) and 3.8 liters of H2O (water).

Answer 1

1 Tbsp = 14.3 g

1) Molar concentration = n(solute) / V(solution) = m(solute)/(Mr(solute)×V(solution)) = (5×14.3) g /(74.44 g/mol × 3.8 L) = 0.253 mol/L

2) Molality = n(solute)/m(solvent) = m(solute)/(Mr(solute)×m(solvent)) = (5×14.3) g /(74.44 g/mol × 3.8 kg) = 0.253 mol/kg

3) % m/m = m(solute)/m(solution) × 100% = (5×14.3) g / ((5×14.3) g + 3800 g) × 100% = 1.85%

4) $\chi$ = n(solute) / n(solution)

n(NaClO) = m(NaClO) / Mr(NaClO) = (5×14.3) g / 74.44 g/mol = 0.96 mol

n(H₂O) = m(H₂O) / Mr(H₂O) = 3800 g / 18 g/mol = 211.11 mol

$\chi$ (NaClO) = 0.96 mol / (0.96 mol + 211.11 mol) = 0.0045

5) % v/v = V(solute) / V(solution) × 100%

V(NaClO) = m(NaClO) / $\rho$ (NaClO) = (5×14.3) g / 1.11 g/mL = 64.4 mL

% v/v = 64.4 mL / 3800 mL × 100% = 1.69%

6) PPM = m(solute)_mg / m(solution)_kg = (5×14.3×10³) mg / 3.8 kg = 18815.8 ppm