How many mili moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 103.57° C between boiling point and freezing point of solution :- ( K f =1.86 K kg mol -1, K b =0.52 K kg mol -1)

Question

How many mili moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 103.57° C between boiling point and freezing point of solution :- ( K f =1.86 K kg mol -1, K b =0.52 K kg mol -1) (A) 500 mili moles (B) 900 milli moles (C) 750 milli moles (D) 650 milli moles

Tardigrade Admin · Accepted Answer

molality =( n / w g ) × 1000 molality =( n /500) × 1000=2 n Δ T b = T b - T b 0 Δ T f = T f 0- T f Δ T b +Δ T f =( T b - T f )+( T f 0- T b 0) ( K b × m + K f × m )=103.57+(0-100) (0.52+1.86) × 2 n =3.57 n =(3.57/2 × 2.38)=(357/2 × 238) mol no. of millimol =(357 × 1000/2 × 238)=750 millimol

Q. How many mili moles of sucrose should be dissolved in 500g of water so as to get a solution which has a difference of 103.57∘C between boiling point and freezing point of solution :- (Kf​=1.86Kkgmol−1,Kb​=0.52Kkgmol−1)

500 mili moles

900 milli moles

750 milli moles

650 milli moles

molality =wg​n​×1000 molality =500n​×1000=2n ΔTb​=Tb​−Tb0​ ΔTf​=Tf0​−Tf​ ΔTb​+ΔTf​=(Tb​−Tf​)+(Tf0​−Tb0​) (Kb​×m+Kf​×m)=103.57+(0−100) (0.52+1.86)×2n=3.57 n=2×2.383.57​=2×238357​mol no. of millimol =2×238357×1000​=750 millimol

Q. How many mili moles of sucrose should be dissolved in $500 g$ of water so as to get a solution which has a difference of $103.5 7^{\circ} C$ between boiling point and freezing point of solution :-
$(K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.52 K k g m o l^{- 1})$