5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82 °C. If Na2SO4 is 81.5%

Molar mass of Na2SO4 = 2(23) + 32 + 4(16) = 142 g/mol
5 g Na2SO4 = 5 g / 142 g/mol = 0.0352 mol
Mass of water = x g = x/1000 kg = 0.001x kg
Molality m = 0.0352 mol Na2SO4 / 0.001x kg H2O = 35.2/x mol/kg
ΔTf = Kf * m * i
where ΔTf is the change in freezing point, Kf is the cryoscopic constant for water (1.86 °C/m), m is the molality, and i is the van't Hoff factor.
For Na2SO4, i = 3 (since it dissociates into 2 Na+ ions and 1 SO42- ion).
3.82 °C = 1.86 °C/m * (35.2/x mol/kg) * 3
3.82 °C = 197.136/x °C
x = 197.136/3.82
x ≈ 51.6 g
The closest option is 45g. However, this solution assumes complete dissociation, which may not be entirely accurate in reality. The discrepancy might also arise from rounding errors or incomplete information in the problem statement (e.g., the actual van't Hoff factor for Na2SO4 at that concentration might differ slightly from 3). Note that the provided data includes 'If Na2SO4 is 81.5?', which is incomplete and doesn't seem to affect the calculation directly, suggesting a missing part of the question.