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Question: Calculate molality of 2.5 grams of Ethanoic Acid \(\left( {C{H_3}COOH} \right)\) in 75 grams of Benz...

Calculate molality of 2.5 grams of Ethanoic Acid (CH3COOH)\left( {C{H_3}COOH} \right) in 75 grams of Benzene.

Explanation

Solution

Hint -Before solving this numerical which is from the basic of physical chemistry we will first define the term molality and then we will proceed by calculating the molar weight of the both the compounds and solve the problem accordingly.

Complete step by step solution:

Molality- It is defined as the number of moles of solute per kilogram of solvent. Its SI unit is mol per kg.
molality = moles of soluteweight of solvent in kg{\text{molality = }}\dfrac{{{\text{moles of solute}}}}{{{\text{weight of solvent in kg}}}}
Given
- The weight of the ethanoic acid (CH3COOH)\left( {C{H_3}COOH} \right) = 2.5gm
- The weight of the benzene = 75gm = 0.075kg
- Now molar weight of the ethanoic acid (CH3COOH)\left( {C{H_3}COOH} \right) =
=12+3×1+12+16×2+1 =60 = 12 + 3 \times 1 + 12 + 16 \times 2 + 1 \\\ = 60 \\\
- Molar weight of benzene C6H6{C_6}{H_6} = = 6 \times 12 + 1 \times 6 \\\ = 78 \\\
From the formula of molality
molality = moles of soluteweight of solvent in kg{\text{molality = }}\dfrac{{{\text{moles of solute}}}}{{{\text{weight of solvent in kg}}}}
Substituting the value of mole of solute and weight of solvent which is ethanoic acid and benzene respectively
molality = 2.560×0.075=0.56mole/kg{\text{molality = }}\dfrac{{2.5}}{{60 \times 0.075}} = 0.56mole/kg

Therefore the molality of the given solutions is 0.56mole/kg0.56mole/kg

Additional Information- A solution is formed, when one substance dissolves into another.

The solution consists of a homogeneous combination of a solvent dissolved into a liquid.
The solution is the dissolved material, while the dissolving medium is the solvent. For several different types of solutes and solvents, solutions can be formed. A solvent is the material which is present in a larger amount in the solution while a solute is the material which is present in the smaller amount.

Note- As we know the concentrations expressed in molality are used for the study of the properties of vapor pressure and temperature changes related solutions and molality is used, as its value does not change with temperature changes. In contrast, the volume of a solution is slightly temperature-dependent.