Question

Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. View Available Hint(s) mNaClm N a C l = nothing nothing Part B Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures. View Available Hint(s) χNaClχ N a C l chi_NaCl = nothing Part C Calculate the concentration of the salt solution in percent by mass. Express your answer to four significant figures and include the appropriate units. View Available Hint(s) percent by mass NaClN a C l = nothing nothing Part D Calculate the concentration of the salt solution in parts per million. Express your answer as an integer to four significant figures and include the appropriate units. View Available Hint(s) parts per million NaClN a C l = nothing nothing

Answer 1

Answer:

m = 2.955x10⁻² mol/kg

X = 5.323x10⁻⁴ mol NaCl/Total moles

(w/w)% = 0.1726%

ppm = 1726 mg/kg

Explanation:

Molality is the ratio between moles of solute per kg of solution.

As the solution is 2.950×10⁻² mol/L, mililters are 999.2mL and density is 0.9982 g/mL, molality is:

m = 2.950×10⁻² mol/L×(1L/0.9982kg) = 2.955x10⁻² mol/kg

Mole fraction is moles of NaCl/total moles.

Moles of H₂O are:

999.2mL×(0.9982g/mL)×(1mol/18,02g) = 55,35 moles of H₂O

Moles of NaCl are:

2.950×10⁻² mol/L×(0.9992L)= 2.950×10⁻² mol of NaCl

mole fraction is:

X = 2.950×10⁻² mol of NaCl / (2.950×10⁻² mol of NaCl+55.35mol water) = 5.323x10⁻⁴ mol NaCl/Total moles

Mass of NaCl is:

2.950×10⁻² mol of NaCl×(58.44g/mol) = 1.724g of NaCl

Mass of water is:

55.35mol water×(18.02g/mol) = 997.4g of H₂O

(w/w)% is:

1.724g of NaCl / (1.724g of NaCl+997.4g of H₂O) ×100 = 0,1726%

Parts per million is mg of NaCl per kg of solution, that is:

1724mg of NaCl / 0.999124g = 1726 ppm

I hope it helps!