Question

8. You prepare a 500 ml solution that contains 436 NOH The solution has a density o has a measured density of 1.100 g/ml The density of 100 ml. and the density o NaOH is 2.13 g/cm. Complete the following table: Mass of Solute Moles of Solute Volume of Solute Mass of Solvent Moles of Solvent m (moles solute/kg solvent) Total Moles of Solution mole fraction NaOH Mass of Solution mole fraction H20 Volume of Solution ppm (mg solute/L solution)

Answer 1

mass of solute = 43.6 g

moles of solute = mass/ molar mass = 43.6 (g) * $\small \frac{1 \: mole }{ 40\: g}$ = 1.09

mass of solvent  

mass of solution = volume of solvent * density = 500 mL * 1.10 (g/mL) = 550 g

mass of solvent =mass of solution - mass of solute = 550 - 43.6 = 506.4 g

moles of solvent (H₂O)  = mass of solvent /molar mass of solvent = 506.4 * $\small \frac{1 \: mole }{ 18\: g}$ = 28.13

total moles of solution = 1.09 +28.13 = 29.22

volume of solution = 500 ml

w/w % = (weight of solute *100 / weight of solution) = (43.6*100/550) = 7.92%

w/v% =   (weight of solute *100 / volume of solution) = (43.6*100/500) = 8.72 %

v/v% = (volume of solute *100/volume of solution)  

now volume of NaOH = mass/ density = 43.6 g *  _{$\small \frac{1 cm^{3}}{2.13 g}$} = 20.469 cm³ = 20.469 mL

so, v/v% = (20.469*100/500) = 4.09 %

M = $\small \frac{moles \: of \: solute *1000}{volume\: of\: solution\: (mL)}$ =  $\small \frac{1.09\times 1000}{500}$ = 2.18 (mol/L)

m = $\small \frac{moles \: of \: solute *1000}{mass\: of\: solvent\: (g)}$ = $\small \frac{1.09\times 1000}{506.4}$ = 2.15 (mol/Kg)

mole fraction of NaOH = $\small \frac{moles \: of \: NaOH}{total\, moles}$ =  $\small \frac{1.09}{29.22}$ = 0.0373

mole fraction of H₂O = 1 - mole fraction of NaOH = ( 1 - 0.0373) = 0.9627

1 ppm = 1 mg/L

hence concentration of NaOH in mg/L

mass of NaOH =43.6 g = 43600 mg

volume of solution = 500 mL = 0.5  L

then , concentration of NaOH = $\small \frac{43600\: mg}{0.5\: L}$ = 87200 mg/L = 87200 ppm