The boiling point of a solution made by dissolving `12.0 g` of glucose in `100 g` of water is `100.34^(@)C`. Calculate the molecular weight of glucose, `K_(b)` for water `= 0.52^(@)C//m`.

Explanation

Using the relation for molecular weight of a solute from elevation in boiling point, we have :
`M_B =K_b ((g_B)/(g_ADeltaT_b)xx1000)`
`rArr M_B=0.52 ((12)/(100xx0.34)xx1000)`
`(DeltaT_b= 100.34-100 =0.34^@C)`
`M_B = 183.5 g //mol`
Depression in Freezing Point :

`DeltaT_f=K_f""(((g_B)/(M_B))/(g_A)xx1000) rArr M_B =K_f""((g_B)/(g_ADeltaT_f)xx1000)`
Osmotic Pressure :
`piV = nRT =(g_B)/(M_B)RT rArr M_B = (g_B RT)/(piV)`