Insulin (C2H10O5)n is dissolved in a suitable solvent and the osmotic pressure (π) of solutions of various concentrations (g/cc) C is measured at 20°C. The slope of the plot of π against ‘C’ is found to be 4.65 × 10-3 The molecular weight of insulin is

Question

Insulin (C2H10O5)n is dissolved in a suitable solvent and the osmotic pressure (π) of solutions of various concentrations (g/cc) C is measured at 20°C. The slope of the plot of π against ‘C’ is found to be 4.65 × 10-3 The molecular weight of insulin is (A) 4.8 × 105 (B) 9 × 105 (C) 3 × 105 (D) 5.17 × 106

Tardigrade Admin · Accepted Answer

π V=(wB/mB) × RT π=((wB/V)) (RT/mB) π=C×(RT/mB)×1000 ldots(i) where C = concentration in g/cc, Comparing eqs. (i) and (ii), ...(ii) Slope = (RT/mR)×1000 mB=(RT/ textSlope)×1000 =(0.0821×293×1000/4.65×10-3) =5.17×106 J

$4.8 \times 1 0^{5}$

$9 \times 1 0^{5}$

$3 \times 1 0^{5}$

$5.17 \times 1 0^{6}$

Q. Insulin (C2​H10​O5​)n​ is dissolved in a suitable solvent and the osmotic pressure (π) of solutions of various concentrations (g/cc) C is measured at 20∘C. The slope of the plot of π against ‘C’ is found to be 4.65×10−3 The molecular weight of insulin is

4.8×105

9×105

3×105

5.17×106

πV=mB​wB​​×RT π=(VwB​​)mB​RT​ π=C×mB​RT​×1000…(i) where C = concentration in g/cc, Comparing eqs. (i) and (ii), ...(ii) Slope =mR​RT​×1000 mB​=SlopeRT​×1000 =4.65×10−30.0821×293×1000​ =5.17×106J

$4.8 \times 1 0^{5}$

$9 \times 1 0^{5}$

$3 \times 1 0^{5}$

$5.17 \times 1 0^{6}$