How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 104- C between boiling point and freezing point- K f -1-86 Kkgmol -1- K b -0-52 Kkgmol -1A- 8-40B- 3-36C- 0-840D- 1-68

The correct option is C 0.840 T_b T_f =104^∘C Δ T_b =T_b T_b^∘ Δ T_f =T_f^∘ T_f (Δ T_b+T_b^∘) (T_f Δ T_f)=104^∘C Δ T_b +100 0+Δ T_f=104 Δ T_b +Δ T_f =4 K_b m ...

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Byju's Answer

Standard XII

Physics

Equipartition Theorem

How many mole...

Question

How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of $104^{\circ} C$ between boiling point and freezing point. $(K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.52 K k g m o l^{- 1})$ .

1.68

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3.36

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8.40

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0.840

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Solution

The correct option is D
0.840

$T_{b} - T_{f} = 104^{\circ} C$

$Δ T_{b} = T_{b} - T_{b}^{\circ}$

$Δ T_{f} = T_{f}^{\circ} - T_{f}$

$(Δ T_{b} + T_{b}^{\circ}) - (T_{f} - Δ T_{f}) = 104^{\circ} C$ \

$Δ T_{b} + 100 - 0 + Δ T_{f} = 104$

$Δ T_{b} + Δ T_{f} = 4$

$K_{b} m + K_{f} m = 4$

$(0.52 + 1.86) m = 4$

$2.38 \times m = 4$

m=1.68

$\frac{m o l e s}{500} \times 1000 = 1.68$

moles $= \frac{1.68}{2} = 0.84$ moles

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How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 104∘C between boiling point and freezing point. (Kf=1.86Kkgmol−1,Kb=0.52Kkgmol−1).

The correct option is D 0.840 Tb−Tf=104∘C ΔTb=Tb−T∘b ΔTf=T∘f−Tf (ΔTb+T∘b)−(Tf−ΔTf)=104∘C\ ΔTb+100−0+ΔTf=104 ΔTb+ΔTf=4 Kbm+Kfm=4 (0.52+1.86)m=4 2.38×m=4 m=1.68 moles500×1000=1.68 moles =1.682=0.84 moles

How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of $104^{\circ} C$ between boiling point and freezing point. $(K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.52 K k g m o l^{- 1})$ .