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Molality : It is defined as the moles of...

Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m.
Molality(m)`=("Number of moles of solute")/("Number of kilograms of the solvent")`
let `w_(A)` grams of the solute of molecular mass `m_(A)` be present in `w_(B)` grams of the solvent, then:
Molality(m) =`(w_(A))/(m_(A)xxw_(B))xx1000`
Relation between mole fraction and molality:
`X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n)`
`(X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A))`
`(X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m`
The mole fraction of the solute in the 12 molal solution of `CaCo_(3)` is :

A

0.822

B

0.177

C

1.77

D

0.0177

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The correct Answer is:
B
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Molality is defined as the number of moles of the solute per kg of the solvent.

Molality : It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted by 'm'. Molality(m)= "Number of moles of solute"/"Number of kilo-grams of the solvent" Let w_A grams of the solute of molecular mass m_A be present in w_a grams of the solvent, then Molality (m)= w_A/(m_Axxw_B)xx100 Realtion between mole fraction and Molality : X_A=n/(N+n) and X_B=N/(N+n) X_A/X_B=n/N="Moles of solute"/"Moles of solvent"=(w_Axxm_B)/(w_Bxxm_A) (X_Axx1000)/(X_Bxxm_B)=(w_Axx1000)/(w_Bxxm_A)=m or (X_Axx1000)/((1-X_A)m_B)=m What is the molality of final solution obtained in the above problem

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  • Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m) =("Number of moles of solute")/("Number of kilograms of the solvent") let w_(A) grams of the solute of molecular mass m_(A) be present in w_(B) grams of the solvent, then: Molality(m) = (w_(A))/(m_(A)xxw_(B))xx1000 Relation between mole fraction and molality: X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n) (X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A)) (X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m If the mole fraction of a solute is changed from (1)/(4) "to" (1)/(2) in the 800 g of solvent then the ratio tof molality will be:

    A
    `1:3`
    B
    `3:1`
    C
    `4:3`
    D
    `1:2`

  • Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m) =("Number of moles of solute")/("Number of kilograms of the solvent") let w_(A) grams of the solute of molecular mass m_(A) be present in w_(B) grams of the solvent, then: Molality(m) = (w_(A))/(m_(A)xxw_(B))xx1000 Relation between mole fraction and molality: X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n) (X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A)) (X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m What is the quantity of water that should be added to 16 g methonal to make the mole fraction of methonal as 0.25?

    A
    27 g
    B
    12 g
    C
    18 g
    D
    36 g

  • Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m) =("Number of moles of solute")/("Number of kilograms of the solvent") let w_(A) grams of the solute of molecular mass m_(A) be present in w_(B) grams of the solvent, then: Molality(m) = (w_(A))/(m_(A)xxw_(B))xx1000 Relation between mole fraction and molality: X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n) (X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A)) (X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m The molality of 1 litre solution with y% by (w/v) pf CaCO_(3) is 2 . The weight of the solvent present in the solution is 900g , then value of y is : [Atomic weight : Ca=40, C=12 , O=16]

    A
    9
    B
    18
    C
    27
    D
    36

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