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Q.At 20 degree C the osmotic pressure of 45 g per litre solution of a substance is 3.2 atmosphere. Calculate the value of solution constant. The molecular weight of the substance is 342.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2019Subjective· 2mImportance★★★★★
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From pi = (w/MV)RT, R = pi/(CT); substituting gives R = 3.2/(0.1316 x 293) = 0.083 L atm K^-1 mol^-1.

Data: pi = 3.2 atm, mass = 45 g in 1 L, M = 342 g mol^-1, T = 20 C = 293 K.

Osmotic pressure: pi = CRT, where C = concentration in mol L^-1.

Step 1 - Molar concentration:

C = (45/342) / 1 L = 0.1316 mol L^-1.

Step 2 - Solve for R (the solution constant):

R = pi / (C x T) = 3.2 / (0.1316 x 293)

R = 3.2 / 38.56

R = 0.0830 L atm K^-1 mol^-1.

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