Question

if mass percent of Glucose in the aqueous solution is 36% find molality of ​

Answer 1

given, mass percent of glucose in the aqueous solution is 36%

in 100g solution, 36g glucose and rest 64g water(solvent )

mole of glucose(solute) = mass of glucose/molar mass of glucose

= 36/180 = 1/5

we know, molality = number of mole of solute × 1000/mass of solvent (in g)

= (1/5) × 1000/64

= 200/64

= 3.125

hence , molarity of glucose is 3.125m