At 300 K. 18g of glucose present per litre its solution has an osmotic pressure of 4.98 bars .If the osmotic pressure of solution is 1.52 bars on the same temperature, what would be its concentration?

Asked by Topperlearning User | 20th Jun, 2016, 03:42: PM

Expert Answer:

    For solution A                          p V= n RT
                                                            4.98 x 1L = 18/180 x R x T
 
               For solution B                         p V= n RT
                                                     1.52 x 1L = n x R x T
                                                1.52 x 1L/ n = 4.98 x 1L / 0.1
                                                 1.52 / n = 4.98 / 0.1
                                                            n = 1.52 x 0.1 / 4.98 = 0.035moles
                                                            c= 0.035 mol L-1

Answered by  | 20th Jun, 2016, 05:42: PM