Home
Class 11
CHEMISTRY
Mole fraction of a solute in an aqueous ...

Mole fraction of a solute in an aqueous solution is 0.2. The molality of the solution will be

A

13.88

B

1.388

C

0.138

D

0.0138

Text Solution

AI Generated Solution

The correct Answer is:
To find the molality of a solution when the mole fraction of the solute is given, we can follow these steps: ### Step 1: Understand the given information We are given the mole fraction of the solute (let's denote it as \(X_{solute}\)) which is 0.2. In an aqueous solution, the solvent is water. ### Step 2: Calculate the mole fraction of the solvent The mole fraction of the solvent (let's denote it as \(X_{solvent}\)) can be calculated using the relationship: \[ X_{solvent} = 1 - X_{solute} \] Substituting the given value: \[ X_{solvent} = 1 - 0.2 = 0.8 \] ### Step 3: Relate mole fractions to moles Let’s assume we have 1 mole of the solution. The moles of solute (\(n_{solute}\)) and moles of solvent (\(n_{solvent}\)) can be expressed as: \[ n_{solute} = X_{solute} \times \text{Total moles} = 0.2 \times 1 = 0.2 \text{ moles} \] \[ n_{solvent} = X_{solvent} \times \text{Total moles} = 0.8 \times 1 = 0.8 \text{ moles} \] ### Step 4: Calculate the mass of the solvent To find the mass of the solvent (water), we use the number of moles of the solvent and the molar mass of water (approximately 18 g/mol): \[ \text{Mass of solvent} = n_{solvent} \times \text{Molar mass of water} = 0.8 \text{ moles} \times 18 \text{ g/mol} = 14.4 \text{ grams} \] ### Step 5: Convert mass of solvent to kilograms Since molality is defined as moles of solute per kilogram of solvent, we convert the mass of the solvent from grams to kilograms: \[ \text{Mass of solvent in kg} = \frac{14.4 \text{ grams}}{1000} = 0.0144 \text{ kg} \] ### Step 6: Calculate molality Molality (\(m\)) is defined as: \[ m = \frac{n_{solute}}{\text{Mass of solvent in kg}} = \frac{0.2 \text{ moles}}{0.0144 \text{ kg}} \approx 13.89 \text{ mol/kg} \] ### Final Answer The molality of the solution is approximately **13.89 mol/kg**. ---

To find the molality of a solution when the mole fraction of the solute is given, we can follow these steps: ### Step 1: Understand the given information We are given the mole fraction of the solute (let's denote it as \(X_{solute}\)) which is 0.2. In an aqueous solution, the solvent is water. ### Step 2: Calculate the mole fraction of the solvent The mole fraction of the solvent (let's denote it as \(X_{solvent}\)) can be calculated using the relationship: \[ ...
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY|Exercise Exercises Assertion-Reasoning|10 Videos
  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY|Exercise Exercises Integers|13 Videos
  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY|Exercise Exercises Multiple Correct (Mole Concept In Solution)|15 Videos
  • S-BLOCK GROUP 2 - ALKALINE EARTH METALS

    CENGAGE CHEMISTRY|Exercise Ex 5.1 Objective|2 Videos
  • STATES OF MATTER

    CENGAGE CHEMISTRY|Exercise Exercises (Ture False)|25 Videos

Similar Questions

Explore conceptually related problems

Mole fraction of urea in its aqueous solution is 0.2, them molality of the solution will be.

Mole fraction of solute in 1 mole aqueous solution is

Mole fraction of a solute in 2.5 molal aqueous solution is

The mole fraction of glucose in aqueous solution is 0.2 then molality of solution will be

Mole fraction of solute in aqueous solution of 30% NaOH.

The mole fraction of glucose in an aqueous solution is 9/109. The molality of solution is:

CENGAGE CHEMISTRY-SOME BASIC CONCEPTS AND MOLE CONCEPT-Exercises Single Correct
  1. The weight of 1 xx 10^(22) molecules of CuSO(4). 5H(2)O is

    Text Solution

    |

  2. How many moles of O(2) will be liberated by one mole of CrO(5) is the ...

    Text Solution

    |

  3. BrO(3)^(ɵ) + 5Br^(ɵ) rarr Br(2) + 3 H(2) O IF 50 mL 0.1 M BrO(3)^(ɵ)...

    Text Solution

    |

  4. To 1 L of 1.0 M impure H(2)SO(4) sample, 1.0 M NaOH solution was added...

    Text Solution

    |

  5. The expression relating mole fraction of solute (chi(2)) and molarity ...

    Text Solution

    |

  6. At 100^(@)C and 1 atm, if the density of the liquid water is 1.0 g cm^...

    Text Solution

    |

  7. Consider the ionisation of H(2) SO(4) as follow" H(2) SO(4) + 2H(2)P...

    Text Solution

    |

  8. Calculate the number of oxygen atoms requried to combine with 7.0 g of...

    Text Solution

    |

  9. 36.5% HCl has density has density equal to 1.20 g mL^(-1). The molarit...

    Text Solution

    |

  10. 10 mL of 1 M BaCl(2) solution and 5 mL 0.5 N K(2)SO(4) are mixed toget...

    Text Solution

    |

  11. Mole fraction of a solute in an aqueous solution is 0.2. The molality ...

    Text Solution

    |

  12. An exess of NaOH was added to 100 mL of a FeCl(3) solution which gives...

    Text Solution

    |

  13. Two samples of HCl of 1.0 M and 0.25 M are mixed. Find volumes of thes...

    Text Solution

    |

  14. If 100 mL of H(2) SO(4) and 100 mL of H(2)O are mixed, the mass percen...

    Text Solution

    |

  15. 12.5 mL of a solution containing 6.0 g of a dibasic acid in 1 L was fo...

    Text Solution

    |

  16. One litre of a sample of hard water contains 5.55 mg of CaCl(1) and 4....

    Text Solution

    |

  17. 10 mL of 0.2 N HCl and 30 mL of 0.1 N HCl to gether exaclty neutralise...

    Text Solution

    |

  18. A metal oxide has the formul Z(2)O(3). It can be reduced by hydrogen t...

    Text Solution

    |

  19. The reaction between yttrium metal and dilute HCl produces H(2) (g) an...

    Text Solution

    |

  20. What volume of H(2) at 273 K and 1 atm will be consumed in obtaining 2...

    Text Solution

    |