Saved Bookmarks
| 1. |
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute ? |
|
Answer» Solution :Vapour pressure of PURE water at the boiling POINT = 1 atm = 1.013 BAR Vapour pressure of solution=1.004 bar Mass of solute = 2 g Mass of solution = 100 g Mass of solvent = 100 - 2 = 98 g Applying Raoult.s LAW, `(p^@ -p_S)/(p^@) = X_2 ~= (w_2 xx m_1) /(m_2 xx W_1) ` `(1.013 -1.004 )/(1.013 ) =(2xx 10 )/(M_2 xx 98 )` `orM_2 =(2xx 18xx 1.013 )/(0.009xx 98 )` |
|