Home
Class 12
CHEMISTRY
What is the percent by mass of iodine ne...

What is the percent by mass of iodine needed to reduce the freezing point of benzene to `3.5^(@)"C"`? The freezing point and cryoscopic constant of pure benzene are `5.5^(@)"C"` and `5.12K/m` respectively.

Text Solution

Verified by Experts

`DeltaT_f=5.2 -3.2 =2^@C`
`DeltaT_f=K_f.m=5.12m rArr m=0.39 =(g_B//M_B)/(g_A//1000)rArr (g_B)/(g_A)=(0.39xx254)/(1000)=0.099~~0.1`
`rArr % "by mass " = (g_B)/(g_A+g_B)=(0.1)/(1+0.1)xx100%=9.09%`
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    VMC MODULES ENGLISH|Exercise PRACTICE EXERCISE -03|4 Videos
  • LIQUID SOLUTIONS

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE -A|10 Videos
  • LIQUID SOLUTIONS

    VMC MODULES ENGLISH|Exercise PRACTICE EXERCISE -01|5 Videos
  • LIQUID SOLUTION

    VMC MODULES ENGLISH|Exercise IMPECCABLE|100 Videos
  • MOCK TEST 1

    VMC MODULES ENGLISH|Exercise PART II : CHEMISTRY (SECTION-2)|10 Videos

Similar Questions

Explore conceptually related problems

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.

A certain solution of benzoic acid in benzene boil at 82.6^(@)C and freezes at 3.1^(@)C . What information about the no of particiles and the structure of benzoic acid at the two temperature can be deduced from the above data ? The boiling point and freezing point of pure benzene are 180.1^(@)C and 5.5^(@)C respectively. K_(r)=5.12 k kg//"mol", k_(b)=2.67 k kg/ mol.

What will be the freezing point of a 0.5 m KCl solution ? The molal freezing point constant of water is 1.86^@C m^(-1) .

A solution containing 0.1 mol of naphthalene and 0.9 mol of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed upto 353 K where its vapour pressure was found to be 670 mm . The freezing point and boiling point of benzene are 278.5 K and 353 K respectively, and its enthalpy of fusion is 10.67 KJ "mol"^(-1) . Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume ideal behaviour.

What mass of NaCl ("molar mass" =58.5g mol^(-1)) be dissolved in 65g of water to lower the freezing point by 7.5^(@)C ? The freezing point depression constant K_(f) , for water is 1.86 K kg mol^(-1) . Assume van't Hoff factor for NaCl is 1.87 .

What mass of NaCI ("molar mass" =58.5g mol^(-1)) be dissolved in 65g of water to tower the freezing point by 7.5^(@)C ? The freezing point depression constant, K_(f) , for water is 1.86 K kg mol^(-1) . Assume van't Hoff factor for NaCI is 1.87 .

What is the molality of solution of a certain solute in a solvent .If there is a freezing point depression of 0.184 ""^(@) C and if the freezing point constant is 18.4 K kg mol^(-1) ?

On adding 1g arsenic to 80 g benzene, the freezing point of benzene is lowered by 0.213^(@)C . The formula of arsenic molecule is : (Atomic mass of arsenic = 75) ( K_(f) = 5.12 K kg/mol)

The freezing point of 0.05 m solution of glucose in water is (K1 = 1.86°C m^(-1) )

Read the following paragraph and answer the question given below: Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02 mol fraction of acetic acid in benzene is 277.4 K. The molal cryoscopic constant of benzene in K "molality"^(-1) is