Question

1. If the outside flask is not dried after vaporizing the liquid, will the unknown's calculated molecular weight be too high or too low? EXPLAIN.

A. Assume the atmospheric pressue is 1 atm but the actual pressue is higher than 1 atm. How will this error affect the calculated molar mass of the unknown? EXPLAIN.

B. If the vapor'svolume were to be incorrectly recorded as 125ml, how will this error affect the calculated molarmass of the unknown? EXPLAIN.

Answer 1

1. If the outside flask is not dried after vaporizing the liquid, will the unknown's calculated molecular weight be too high or too low? EXPLAIN.

Solution :- If the flask is not dried after evaporating the liquid from the flask and we used it to record the mass of the sample then moisture present inn it will give the higher weight of the sample than actual present. Molar mass is nothing but the mass per mole therefore the moisture will increase the molar mass too high.

A. Assume the atmospheric pressue is 1 atm but the actual pressue is higher than 1 atm. How will this error affect the calculated molar mass of the unknown? EXPLAIN.

Solution :- If the pressure inside is higher than the atmospheric pressure then the calculated molar mass of the sample is higher than the actual molar mass

Because, n=PV/RT pressure is present at numerator therefore if the value of the P is considered lower than actual then the moles of the gas would be lower so the molar mass of the sample would be higher than actual.

B. If the vapor'svolume were to be incorrectly recorded as 125ml, how will this error affect the calculated molarmass of the unknown? EXPLAIN

Solution :- If the volume of the vapor is recorded incorrectly then the mole calculation will be incorrect this lead to the incorrect molar of the sample

If the 125 ml is the higher than the actual volume of the vapor then molar mass would be lower and if the 125 ml is less than the actual volume then molar mass would be higher.