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Question: What is the molality of \( 5.0g \) mass of toluene dissolved in 255g of benzene?...

What is the molality of 5.0g5.0g mass of toluene dissolved in 255g of benzene?

Explanation

Solution

We know that molality is defined as the ratio of moles of solute to the mass of solvent in kilograms. So, to solve this question, first determine the number of moles corresponding to 5.0g5.0g mass of toluene and then substitute the given values in the formula of molality to get the final answer.
Molality m=moles of solutemass of solvent (kg)m = \dfrac{{{\text{moles of solute}}}}{{{\text{mass of solvent (kg)}}}}

Complete answer:
Toluene is an aromatic hydrocarbon which is composed of a benzene ring linked to one methyl group. It is represented by the chemical formula C6H5CH3{C_6}{H_5}C{H_3} . The atomic mass of carbon and hydrogen is 12gmol−112gmo{l^{ - 1}} and 1gmol−11gmo{l^{ - 1}} respectively. So, the molar mass of toluene is as follows:
Mtoluene=12×6+5×1+12×1+3×1{M_{toluene}} = 12 \times 6 + 5 \times 1 + 12 \times 1 + 3 \times 1
⇒Mtoluene=92.14gmol−1\Rightarrow {M_{toluene}} = 92.14gmo{l^{ - 1}}
Thus, the molar mass of toluene is 92.14gmol−192.14gmo{l^{ - 1}} .
The number of moles of a substance is equal to the ratio of given mass of the substance to its molar mass and as per given question, the mass of toluene is 5.0g5.0g . Therefore, the number of moles of toluene are as follows:
n=given massmolar massn = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}
Substituting values:
⇒n=592.14\Rightarrow n = \dfrac{5}{{92.14}}
⇒n=0.054 moles\Rightarrow n = 0.054{\text{ moles}}
Now, the molality of the solution can be calculated as follows:
m=moles of solutemass of solvent (kg)m = \dfrac{{{\text{moles of solute}}}}{{{\text{mass of solvent (kg)}}}}
As per our calculations and given values, the number of moles of solute i.e., toluene is 0.0540.054 moles and the mass of solvent i.e., benzene is 255g255g . Substituting values in the given formula for molality:
⇒m=0.054255×10−3                    [∵1g=10−3kg]\Rightarrow m = \dfrac{{0.054}}{{255 \times {{10}^{ - 3}}}}\;\;\;\;\;\;\;\;\;\;\left[ {\because 1g = {{10}^{ - 3}}kg} \right]
⇒m=0.212  molkg−1\Rightarrow m = 0.212\;{\text{molk}}{{\text{g}}^{ - 1}}
Hence, the molality of the given solution is 0.212  molkg−10.212\;{\text{molk}}{{\text{g}}^{ - 1}} .

Note:
Remember to check the unit of mass of solvent before substituting it in the formula as the mass of solvent is always to be taken in kg. Also, it is important to note that the molality is always preferred over molarity because it is independent of volume and thus, it is independent of the value of temperature as well.