The vapour pressure of water at 23^ C is 19.8 ~mm , 0.1 mole of glucose is dissolved in 178.2 ~g of water. What is the vapour pressure (in mm ) of the resultant solution?

Step-by-step solution

Given P_s=19.8 ~mm aligned & n_A=0.1 & n_B= 178.2 18 =9.9 aligned According to Raoult's law gathered P_s-P P_s = n_A n_A+n_B 19.8-P 19.8 = 0.1 9.9+0.1 gathered or aligned 198-10 P & =19.8 0.1 10 P & =198-1.98 10 P & =196.02 P & =19.602 ~mm aligned