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InterviewSolution
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10 g of a substance were dissolved in water and the solution was made up to 250 cm^(3). The osmotic pressure of the solution was found to be 8xx10^(5)Nm^(-2) (pascals) at 288 K. Find the molar mass of the solute. |
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Answer» Solution :Here, `w_(2)=10g, V=250cm^(3)=250xx10^(-6)m^(3)=2.5xx10^(-4)m^(3)` `pi=8xx10^(5)Mn^(-2),R="8.314 J K"^(-1)"mol"^(-1), T = 288K` `THEREFORE""M_(2)=(w_(2)RT)/(piV)=((10g)("8.314 Nm K"^(-1)"mol"^(-1))(288K))/((8xx10^(5)Nm^(-2))XX(2.5xx10^(-4)m^(3)))(1J = 1Nm)="119.7 G mol"^(-1)` |
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