Question

consider naphthalene,C10H8(MM=128.2 g/mol), the active ingredient of some moth balls. A solution of naphthalene is prepared by mixing 35 g of naphthalene with 1.2 L of carbon sulfide, CS2 (d=1.263 g/mL). assume the volume remains 1.20 L when the solution is prepared.

a)what is the mass percent of naphthalene in the solution?

b) represent the concentration of naphthalene in parts per million.

c) what is the density of the solution?

d)what is the molarity of the solution?

e)what is the molality of the solution?

f) what is the volume of the solution is required to prepare 100 mL of 0.12 M solution of naphthalene in CS2?

Answer 1

a) first que find the mass of the carbon sulfide:

m= dxv m = (1,263 g/mL) (1200 mL) = 1515.6 g of carbon sulfide

now the entire mass of the solution will be 35 g + 1515.6 g = 1550.6 g

% m/m = 35g (naphtalene) / 1550.6 g (total solution) x 100 =2.25 m/m% of naphtalene

b) 1ppm = 0.0001%

then we do the relation.

naphtalen ppm= 2.25%(1ppm) / 0.0001% = 22500 ppm of naphtalene.

c) to find the density we divide the entire mas per the mL of solution

d = 1550.6 g / 1200 mL = 1.29 g/mL

d)we find the moles of naphtalene dividin pero the MW of the naphtalene

n = (35g)/ 126.2g/mol) = 0.27 mol of naphtalene

M= 0.27 mol/ 1.2 L = 0.23 M

e) We transform the gr we found of the entire solution to Kg

b = moles of nafthalene / Kg of solution

b = 0.27 mol/ 1.550 Kg = 0.17 mol/Kg

f) M₁V₁ = M₂V₂   this is because the moles remind the same

then V₂= [(100 mL) (0.12 M)]/0.23 M V₂= 52.2 mL