x g of urea y g of glucose r dissolved in 1000g of water in a second experiment y - Chemistry - Solutions

Question

x g of urea y g of glucose r dissolved in 1000g of water in a
second experiment y g of urea x g of glucose are dissolved in 1000
g of water the depression of freezing point in the first case is
twice the depression in the second case calculate the ratio x:y

Kumar Sarang · Accepted Answer

Solution 1: no of moles of urea = x/60 moles no of moles of glucose = y/180 moles molality of solution = no of moles of solute / mass of solvent(in kg) = ( (x/60)+(y/180) )/1 mol /kg from the relation $Δ T_{b} = K_{b} m$ we get , $Δ T_{b}$ = $K_{b}$ * molalitu of solution = $K_{b} * (3 x + y) / 180 (1)$ equation 1 Solution 2 : no of moles of urea = y/60 moles no of moles of glucose = x/180 moles molality of solution = no of moles of solute / mass of solvent(in kg) = ( (y/60)+(x/180) )/1 mol /kg from the relation $Δ T_{b} = K_{b} m$ we get , $Δ {T_{b}}^{\sim}$ = $K_{b}$ * molalitu of solution = $K_{b} * (3 y + x) / 180 (1)$ equation 2 now according to question $Δ T_{b} = 2 Δ {T_{b}}^{\sim}$ Therefore equating equation 1 and 2 we get $K_{b} * (3 x + y) / 180 (1)$ =2 $K_{b} * (3 y + x) / 180 (1)$ thus 3x+y = 6y + 2x x=5y thus x:y = 5:1

x g of urea y g of glucose r dissolved in 1000g of water. in a second experiment y g of urea x g of glucose are dissolved in 1000 g of water.the depression of freezing point in the first case is twice the depression in the second case.calculate the ratio x:y.