Question

10 gm compound dissolved in 0.200 kg of benzene boils at 81.2° c calculate the the molecule weight of compound Kb of benzene =2.53°c kg mol ¹ B.pof benzene =80.1°c​

Answer 1

Let us use some known facts. Let the given differential equation be

Factorize into linear factors. Then use the following results:

Corresponding to each non-repeated factor , the part of C.F. is taken as

In our case,

0 STEP: We factor the expression

Conclusion,

1 STEP: Let find C.F.

Then,

2 STEP: Let find P.I.

Conclusion,

Answer 2

Answer:

Answer

AnswerOpen in answr app

AnswerOpen in answr appLet W g of compound is to be dissolved.

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256W

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kg

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solute

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2Wm

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2WmThe depression in the freezing point, ΔTf=0.48K

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2WmThe depression in the freezing point, ΔTf=0.48KThe molal depression in freezing point constant =Kf=5.12Kkg/mol

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2WmThe depression in the freezing point, ΔTf=0.48KThe molal depression in freezing point constant =Kf=5.12Kkg/molΔTf=Kfm

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2WmThe depression in the freezing point, ΔTf=0.48KThe molal depression in freezing point constant =Kf=5.12Kkg/molΔTf=Kfm0.48=5.12×19.2W

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2WmThe depression in the freezing point, ΔTf=0.48KThe molal depression in freezing point constant =Kf=5.12Kkg/molΔTf=Kfm0.48=5.12×19.2WW=5.120.48×19.2=1.8

AnswerOpen in answr appLet W g of compound is to be dissolved.Number of moles of compound, n=molar massmass=256WMass of benzene = 75 g =100075kg=0.075kgMolality of solution, m=mass of solvent in kgnumber of moles of solutem=256×0.075W=19.2WmThe depression in the freezing point, ΔTf=0.48KThe molal depression in freezing point constant =Kf=5.12Kkg/molΔTf=Kfm0.48=5.12×19.2WW=5.120.48×19.2=1.8Hence, 1.8 g of solute is to be dissolved.