5 g of Nacl is dissolved in 1000 g of water if the density of the resulting solution is 0.997 g per ml calculate the molality ,molarity , normality, & mole fraction of the solute assuming volume of solution is equal to that of solvent
Answers
Answer:
molecular mass of NaCl=23*1+35.5*1
=58.5g/mol
molality=mass of solute*1000/molecular mass*mass of solvent(in g)
=5*1000/58.5*1000
.0854m
weight of solution=weight of solvent+weight of solute
=1000+5
=1005g
1 liter=1000ml=1cm cube
density=mass/volume
.997=1005/volume
volume=1005/.997
=1008.02 l
molarity=mass of solute in g/molecular mass*volume of solution in l
=5/58.5*1008.02
=5/58969.17
=8.479M
Explanation:
Given info : 5 g of Nacl is dissolved in 1000 g of water and the density of the resulting solution is 0.997 g/ml.
To find :
- molality
- molarity
- normality
- mole fraction
solution : mass of solution = mass of solute + mass of solvent
here, mass of solute (NaCl) = 5g
mass of solvent ( water ) = 1000g
now the mass of solution = 5g + 1000g = 1005g
∴ volume of solution = mass of solution/density of solution
= = 1008.024 ml
no of moles of solute = mass of solute/molar mass of solute
= = 0.0855 mol
∴ molarity = no of moles of solutel/volume of solution in L
= = 0.0848 M
therefore the molarity of the solution is 0.0848M.
molality of the solution = no of moles of solute/mass of solvent in Kg
= = 0.0855 molal
therefore the molality of the solution is 0.0855molal.
normality of the solution = n - factor × molarity
here NaCl dissociates into Na⁺ and Cl⁻.
∴ n - factor = change in oxidation state of Na⁺/Cl⁻ = 1
now the normality of the solution = 1 × 0.0848 = 0.0848 N
no of moles of solvent( water) , n₁ = mass of water/molar mass of water
= = 55.5 mol
no of moles of solute (NaCl ) , n₂ = 0.0855 mol
∴ mole fraction of solute =
=
therefore the mole fraction of the solute is 0.00154