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प्रश्न
A solution contains 0.8960 g K2SO4 in 500 mL solution. Its osmotic pressure is found to be 0.690 atm at 27°C. Calculate the value of van’t Hoff factor.
संख्यात्मक
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उत्तर
Given: Mass of K2SO4 = 0.8960 g
Volume of solution (V) = 500 mL = 0.5
Temperature (T) = 27°C = 300 K
Observed osmotic pressure `pi_"obs"` = 0.690 atm
Molar mass of K2SO4 = 39.1 × 2 + 32.1 + 16 × 4 = 174.3 g/mol
Gas constant (R) = 0.0821 J K−1 mol−1
Moles of K2SO4 = `0.8960/174.3`
= 0.00514 mol
Let no. of dissociation (i) = 1
`pi_"theory" = CRT`
= `0.00514/0.5 xx 0.0821 xx 300`
= 0.01028 × 24.63
= 0.253 atm
Van’t Hoff factor (i) = `pi_"obs"/pi_"theory"`
= `0.690/0.253`
= 2.72
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पाठ 2: Solutions - REVIEW EXERCISES [पृष्ठ १०५]
