In the standardization of the Na2S2O3 solution, if you did not titrate the liberated iodine immediately, how would this likely affect the value of the molarity you calculate?

The titration should be done fastly because if not titrated immediatly the liberated I2 will go out of the system. So it will not react with the Na2S2O3. then the value we are getting for the Na2S2O3 is not the actuall value. The malority will smaller than the actual value.