# What is the molecular weight of the unknown substance?

a 17.5 g sample of unknown substance was dissolved in 100. g of water. the boiling point of this solution was 100.69 C. What is the molecular weight of the unknown substance? Assume that the unknown does not dissociate. The boiling point elevation constant for water is 0.512 (kg solvent*C)/(mol solute).

Relevance

deltaTb =i x kb xm

change in temperature= 100.69C -100.0C =6.69C

0.69 = (0.512)(m)

13.06 = m

moles of solute = (1.347 x0.100 kgH20) =0.1347 moles

molecular weight= 17.5g/ 0.1347 =130g

Source(s): college chemistry student
• given that:

mass of substance= 17.5 grams

mass of H2O = 100 grams

Tt ( temperature boiling solution) = 100.69 C

To ( temperature boiling water) = 100 C (been noted thou)

Kb(boiling point constant) = 0.512 (kg solvent*C)/(mol solute)

the formula is :

ΔTb = Kb x (mass of substance/Mr ) x (1000/Mass of H2O)

(Tt - To) = Kb x (mass of substance/Mr ) x (1000/Mass of H2O)

(100-100.69) = 0.512 x 17.5/Mr x1000/100

Mr= 129.86 ---> 130