What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?

1 answer:

3 0

The molality of sodium chloride comes to 2.55 mol/kg.
V(solution) = 100 ml.
The mass of the solution (m) is calculated using the formula: m(solution) = d(solution) · V(solution).
Here, m(solution) would be 1.10 g/ml multiplied by 100 ml.
Thus, m(solution) equals 110 g.
The mass fraction of NaCl (ω(NaCl)) is 13.0% or 0.13.
Substituting this into the formula gives m(NaCl) = ω(NaCl) · m(solution).
This results in m(NaCl) equaling 0.13 multiplied by 110 g.
Concluding with m(NaCl) equals 14.3 g.
The number of moles of NaCl (n(NaCl)) is calculated by n(NaCl) = m(NaCl) ÷ M(NaCl).
So, n(NaCl) equals 14.3 g divided by 58.5 g/mol.
This results in n(NaCl) = 0.244 mol.
The mass of water (m(H₂O)) is found as 110 g minus 14.3 g.
Thus, m(H₂O) equals 95.7 g which converts to 0.0957 kg.
Finally, b(NaCl) is calculated as n(NaCl) ÷ m(H₂O).
Hence, b(NaCl) = 0.244 mol divided by 0.0957 kg.
This yields b(NaCl) = 2.55 mol/kg.

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