Hello,
Weight of the unknown liquid inside the bulb after completion of reaction is 50.1879 grams - 50.0230 grams =0.1649 grams.
Volume of the water= 102.7535 grams - 50.0230 grams = 52.7305 grams = volume of the vapour of liquid after its immersion in oil bath at 423.15 K( 150° C) and 0.980 bar.
Using ideal gas law equation PV=nRT, [itex]\frac{0.980 bar*0.0527305 liters}{0.0831447 bar*liters/(mol*K)*423.15 K}=0.001468787 mol [/itex]
Molar weight of the unknown liquid =0.1649 grams/0.001468787 mol= 112.269 grams/mol. Its molecular weight is 1.86427e-22 u.