Question

Determine the boiling point of 1M solution of sugar,glucose,sodium chloride,barium chloride,aluminium chloride

Answer 1

The solvent is not specified. I suppose it is water.

Elevation in boiling point is given by Clausius-Clapeyron relation and Raoult's law combined together. That is :

$\Delta T_b= K_b\ b_{solute}\ i$

Where the Ebullioscopic Constant of solvent is K_b
$K_b=\frac{R\ T_b^2 M}{\Delta H_v}\\\\\Delta H_v=Latent\ heat\ of\ vaporization.\\T_b=Boiling\ point.\\M=Molality\\R=Universal\ Gas\ Constant$
For water K_b = 0.512 °C/Molal

b_solute = Molality of solute = 1 M  given
i = number of particles per molecule of solute after dissociation into ions

i = 1 for Glucose and Sugar , as they do not dissociate in water as ions
  = 1.9 for Na Cl    or 2 theoretically 
  = 3  for Ba Cl2   theoretically
  = 4  for Al Cl3   theoretically.

So Elevations in boiling points are :
 = 0.512 * 1 * 1  = 0.512° C  for  Sugar and Glucose
 = 0.512 * 1 * 2 =  1.024 °C  for Salt   theoretically
 = 0.512 * 1 * 3 = 1.536 °C for Barium Chloride
 = 0.512 * 1 * 4 = 2.048°C  for Aluminum Chloride