Answer to Question #69709 in Inorganic Chemistry for diksha tomar

Question #69709
the mole fraction of 1% (w/v) NaOH sol. is , if density of the solution is 1.2 gm/ml
1
Expert's answer
2017-08-11T06:30:26-0400
Solution:
1 % (w/v) means 1 gram (weight) of NaOH in 100 mL (volume) of solution
n(NaOH) = m(NaOH)/M(NaOH) = 1 g/(23 + 16 + 1) g/mol = 1 g / 40 g/mol = 0.025 mol
m(solution) = ρ×V = 1.2 g/mL × 100 mL = 120 g
m(H2O) = m(solution) – m(NaOH) = 120 – 1 = 119 g
n(H2O) = m(H2O)/M(H2O) = 119 g / 18 g/mol = 6.6 mol
n(solution) = n(H2O) + n(NaOH) = 6.6 + 0.025 = 6.625 mol
XNaOH = 0.025 / 6.625 = 0.004

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