Home
Class 12
CHEMISTRY
Pure water freezes at 273 K and 1 bar. T...

Pure water freezes at `273 K` and 1 bar. The addition of `34.5 g` of ethanol to `500 g` of water changes the freezing point of the solution. Use the freezing point depression contant of water as `2 K kg mol^(-1)`. The figures shown below represent polts of vapour (V.P.) versus temperature (T). [Molecular weight of ethanol is `46g mol^(-1)`]
Among the following, the option representing change in the freezing point is

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
A
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SOLUTIONS AND COLLIGATIVE PROPERTIES

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Test|4 Videos

  • SOLUTIONS AND COLLIGATIVE PROPERTIES

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Passage|2 Videos

  • SOLUTIONS AND COLLIGATIVE PROPERTIES

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Objective Questions II|1 Videos

  • SOLID STATE

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Passage|1 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Subjective Questions|2 Videos

Similar Questions

Explore conceptually related problems

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2K kg "mol"^(-1) . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g "mol"^(-1) ]. Among the following, the option representing change in the freezing point is :

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.

Knowledge Check

  • Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol^(-1). The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol^(-1) Among the following, the option representing change in the freezing point is

    A
    B
    C
    D

  • Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water at "2 K kg mol"^(-1) . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T) [Molecular weight of ethanol is "46 g mol"^(-1) ]. Among in the freezing point is

    A
    B
    C
    D

  • Pure water freezes at 273 K and 1 bar. The addition of 34.5g of ethanol to 500g of water changes the freeing point of the solution. Use of f.p. depression constant of water as 2Kg mol^(-1) . The figures shown below represents plot of vapour pressure (V.P.) versus temperature (T) (mol wt. of ethanol =46 g mol^(-1) ). Among the following the option representing change in the f.p. is

    A
    B
    C
    D

    • Similar Questions

    Explore conceptually related problems

    How much amount of NaCl should be added to 600 g fo water (varphi =1.00 g//mL) to decrease freezing point of water to -0.2^(@)C ? ___________________. (The freezing point depression constant for water =2K kg "mol"^(-1) )

    By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^(@)C . Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol^(-1))

    How much amount of NaCl solution be added to 600 g of water (rho=1.00 g/mL) to decrease the freezing point of water to -0.2^(@)C ? (The freezing point depression constant for water =2 K kg mol^(-1) )

    45 g of ethylene glycol (C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezing point of the solution is (K_(f) for water is 1.86 K kg mol^(-1) )

    A solution containing 8.0 g of nicotine in 92 g of water-frezes 0.925 degrees below the normal freezing point of water. If the freezing point depression constant K_(f) = -1.85^(@)C mol^(-1) then the molar mass of nicotine is -

    Home
    Profile