Question

Calculate the molar mass of a compound, when $6.21g$ of the compound is dissolved in $24g$ of chloroform. The boiling point of pure chloroform is ${68.04^o}C$ and the boiling point of solution is ${61.7^o}C$ , ${k_b}$ of chloroform is 3.63 $Kkg{\text{ }}mo{l^{ - 1}}$

Answer 1

When we dissolve a non-volatile compound in a solvent, its boiling point increases. Boiling point of the solution is always higher than the boiling point of pure solvent. The increase in boiling point is directly proportional to the molality of the solute in the solvent. ${k_b}$is the proportionality constant.

Complete step by step solution:
We know that boiling point of a solution is defined as the temperature at which the vapour pressure of the solution becomes equal to atmospheric pressure. When we dissolve a non-volatile compound in a solvent, the vapour pressure of the solvent decreases and thus the boiling point increases. Therefore when we dissolve a compound in chloroform, its boiling point increases.
$T_b^o =$Boiling point of pure chloroform $=$ ${61.7^o}C$
${T_b} =$ Boiling point of solution $=$ ${68.04^o}C$
Convert the temperatures into Kelvin
$T_b^o =$ $61.7 + {\text{273}}{\text{.15 = 334}}{\text{.85}}K$
${T_b} =$ $68.04 + 273.15 = 341.19K$
Calculate the change in boiling point
$\Delta {T_b} = {T_b} - T_b^o$
$\Rightarrow 341.19K - 334.85K \\\ \Rightarrow 6.34K \\\ \\\$
Number of moles of solute is the ratio of given mass of solute and molar mass of solute.
Number of moles of solute $=$ $\dfrac{{{w_2}}}{{{M_2}}}$ $= \dfrac{{6.21}}{{{M_2}}}$
${w_2} =$ Given mass of solute $=$ $6.21g$
${M_2}$ = Molar mass of solute which has to be calculated
Molality is defined as the number of moles of solute that is a compound dissolved in 1 $kg$ of solvent that is chloroform.
$m =$ Molality $=$ $\dfrac{{{w_2} \times 1000}}{{{M_2} \times {w_1}}}$
${w_1} =$Mass of solvent that is chloroform in $K{\text{ }}kg{\text{ }}mo{l^{ - 1}}$ = $\dfrac{{24}}{{1000}}kg$
Substitute the values in the molality formula
$m =$ $\dfrac{{{w_2} \times 1000}}{{{M_2} \times {w_1}}}$ $= \dfrac{{6.21 \times 1000}}{{{M_2} \times 24}}$
${k_b}$ is boiling point elevation constant and is measured in $Kkg{\text{ }}mo{l^{ - 1}}$.
${k_b}$ $=$ 3.63$$$Kkg{\text{ }}mo{l^{ - 1}}$$ Put values in the formula \Delta {T_b} = {k_b} \times m$$ \Rightarrow 6.34K = \dfrac{{6.21g \times 1000 \times 3.63Kkgmo{l^{ - 1}}}}{{{M_2} \times 24kg}}$$ \Rightarrow {M_2} = \dfrac{{6.21g \times 1000 \times 3.63Kkgmo{l^{ - 1}}}}{{6.34K \times 24kg}}$Solving the above equation, we get$ \Rightarrow {M_2} = 148.3gmo{l^{ - 1}}$

Hence, the Molar mass of the compound is $148.3gmo{l^{ - 1}}$.

Note: When reading the question, we should note the units of measurements. For example it is necessary to convert $24g$ of chloroform into $24kg$ of chloroform. It is also important to remember to convert units of temperature, degree Celsius should be converted into Kelvin.