The osmotic pressure due to 2.20 g polyethylene (I=1) dissolved in enough benzen

Question

The osmotic pressure due to 2.20 g polyethylene (I=1) dissolved in enough benzene to make 100 ml of solution is 1.10 x 10^-2 atm at 25C. Calculate the molar mass of polyethylene. Would the osmotic pressure be higher or lower than 1.10 x 10^-2 atm if the solution was made with 2.20 g of a polyethylene that has a greater average molar mass? Explain. The osmotic pressure due to 2.20 g polyethylene (I=1) dissolved in enough benzene to make 100 ml of solution is 1.10 x 10^-2 atm at 25C. Calculate the molar mass of polyethylene. Would the osmotic pressure be higher or lower than 1.10 x 10^-2 atm if the solution was made with 2.20 g of a polyethylene that has a greater average molar mass? Explain. Would the osmotic pressure be higher or lower than 1.10 x 10^-2 atm if the solution was made with 2.20 g of a polyethylene that has a greater average molar mass? Explain.

Explanation / Answer

Answer: Vant Hoff's equation V = nRT

Given =0.011atm, V=100ml, R=0.0821 L atm mol-1, T=25deg C=298K and wt of polyethylene=2.20g/100ml=22g/L

n=wt/mwt

Finally M.wt=(wt*RT)/V= (22*0.0821*298)/(0.011)=48.93g =4893 1.6 g/mol

The osmotic pressure is constant when the solution made with 2020g of polyethylene that has a greater average molar mass described in vant Hoff's equation.

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