JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(5\, g\) of \(Na_2SO_4\) was dissolved in \(x\,g\) of \(H_2O\). The change in freezing point was found to be \(3.82\,^oC\). lf \(Na_2SO_4\) is \(81.5\%\) ionised, the value of \(x\) (\(K_f\) for water \(= 1.86\,^oC\, kg\, mol^{-1}\)) is approximately .............. \(\mathrm{g}\) (molar mass of \(S = 32\, g\, mol^{-1}\) and that of \(Na = 23\, g\, mol^{-1}\))
- A \(15\)
- B \(25\)
- C \(45\)
- D \(65\)
Answer & Solution
Correct Answer
(C) \(45\)
Step-by-step Solution
Detailed explanation
Molality (experimental) \( = \,\frac{{\Delta {T_f}}}{{{K_f}}}\, = \,\frac{{3.82}}{{1.86}}\, = \,2.054\,mol/1000\,g\) solvent Molality (theoretical) \( = \,\frac{{mole\,\,of\,solute}}{{wt.\,of\,solventing\,(g)}} \times 1000\)…
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