Home
Class 12
CHEMISTRY
The freezing point of the H2O is :...

The freezing point of the `H_2O` is :

A

`273.15 K`

B

`268.5K`

C

`234.2K`

D

`150.9K`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the freezing point of water (H₂O), we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Concept of Freezing Point**: - The freezing point is the temperature at which a substance changes from a liquid to a solid. For water, this is the temperature at which it begins to freeze. 2. **Identify the Freezing Point of Water**: - The freezing point of water is commonly known to be 0 degrees Celsius (°C). 3. **Convert Celsius to Kelvin**: - To convert Celsius to Kelvin, we use the formula: \[ K = °C + 273.15 \] - Substituting the freezing point of water: \[ K = 0 + 273.15 = 273.15 \, K \] 4. **Conclusion**: - Therefore, the freezing point of water (H₂O) is 273.15 Kelvin (K). ### Final Answer: The freezing point of H₂O is **273.15 K**. ---

To determine the freezing point of water (H₂O), we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Concept of Freezing Point**: - The freezing point is the temperature at which a substance changes from a liquid to a solid. For water, this is the temperature at which it begins to freeze. 2. **Identify the Freezing Point of Water**: ...
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    RESONANCE ENGLISH|Exercise EXERCISE-3(PART-2)|16 Videos
  • SOLUTIONS

    RESONANCE ENGLISH|Exercise EXERCISE-3(PART-3)|30 Videos
  • SOLUTIONS

    RESONANCE ENGLISH|Exercise EXERCISE-2(PART-4)|4 Videos
  • SOLUTION AND COLLIGATIVE PROPERTIES

    RESONANCE ENGLISH|Exercise PHYSICAL CHEMITRY (SOLUTION & COLLIGATIVE PROPERTIES)|52 Videos
  • STEREOISOMERISM

    RESONANCE ENGLISH|Exercise EXERCISE (PART III : PRACTICE TEST-2 (IIT-JEE (ADVANCED PATTERN))|23 Videos

Similar Questions

Explore conceptually related problems

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) )

0.1 mole of sugar is dissolved in 250 g of water. The freezing point of the solution is [K_(f) "for" H_(2)O = 1.86^(@)C "molal"^(-1)]

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.

30mL of CH_(3)OH (d = 0.780 g cm^(-3)) and 70 mL of H_(2)O (d = 0.9984 g cm^(-3)) are mixed at 25^(@)C to form a solution of density 0.9575 g cm^(-3) . Calculate the freezing point of the solution. K_(f)(H_(2)O) is 1.86 kg mol^(-1)K . Also calculate its molarity.

The freezing point of nitrobenzene is 3^(@)C . When 1.2 g of chloroform (mol. Wt. =120) is dissolved in 100 g of nitrobenzene, freezing point will be 2.3^(@)C . When 0.6 g of acetic acid is dissolved in 100 g of nitrobenzene, freezing point of solution is 2.64^(@)C . If the formula of acetic acid is (CH_(2)O)_(n) , find the value of n.

When HgI_(2) is added in KI solution. The freezing point of solution:-

0.01 M solution of KCl and CaCl_(2) are separately prepared in water. The freezing point of KCl is found to be -2^(@)C . What is the freezing point of CaCl_(2) aq. Solution if it is completely ionized?

The freezing point of a solution that contains 10 g urea in 100 g water is ( K_1 for H_2O = 1.86°C m ^(-1) )

The freezing point of a solution of 2.40 g of biphenyl ( C_(12)H_(10) ) in 75.0 g of benzene ( C_(6)H_(6) ) is 4.40^(@)C . The normal freezing point of benzene is 5.50^(@)C . What is the molal freezing point constant (@C//m) for benzene ?

4.0g of substance A dissolved in 100g H_(2)O depressed the freezing point of water by 0.1^(@)C while 4.0g of another substance B depressed the freezing point by 0.2^(@)C . Which one has higher molecular mass and what is the relation ?