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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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To find the mole fraction of the solute in a 12 molal solution of CaCO₃, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Definitions**: - Molality (m) is defined as the number of moles of solute per kilogram of solvent. - The formula for molality is given by: \[ ...
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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:

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?

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