the b.pt of pure acetone is 56.38 degree c when 0.707 g of a compound is dissolved in 10 g of acetone there is elevation to 56.88 degree c in bpt what is the molar wt of compound??
Dear student,
Elevation in boiling point=56.88-56.38=0.5 C
as we know Tb=mKb
Where Tb is the elevation in boiling point, m being the molality and Kb being the constant which is not mentioned in the question.
Molality is moles of solute in 1000 gram of solvent
m=(w*1000)/(M*W)
W is weight of solvent, w being the weight of the solute, M is the molar mass of the solvent.
Substituting this in the above equation we will get the molar mass of the solute.
As value of Kb is missing so I am unable to fully solve it .
Tb={(w*1000)/(M*W)}*Kb
Regards
Elevation in boiling point=56.88-56.38=0.5 C
as we know Tb=mKb
Where Tb is the elevation in boiling point, m being the molality and Kb being the constant which is not mentioned in the question.
Molality is moles of solute in 1000 gram of solvent
m=(w*1000)/(M*W)
W is weight of solvent, w being the weight of the solute, M is the molar mass of the solvent.
Substituting this in the above equation we will get the molar mass of the solute.
As value of Kb is missing so I am unable to fully solve it .
Tb={(w*1000)/(M*W)}*Kb
Regards