A solution of $0.5126g $ of naphthalene (molar mass= $128.17gmo{l^{ - 1}}$) in $50.00g$ of $CC{l_4}$ gives a boiling point elevation of $0.402K$ . While a solution of $0.6216g$ of unknown solute in the same mass of the solvent gives a boiling point elevation of $0.642K$.Find the molar mass of the unknown solute.(${K_b}$ for $CC{l_4} = 5.03Kkgmo{l^{ - 1}}$ of solvent).
Answer
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Hint: This is based on the colligative property i.e. Elevation in boiling point. Here naphthalene is solute and \[CC{l_4}\] is solvent. We have to apply formula of elevation in boiling point which is directly proportional to the molal concentration of solute i.e.
$\Delta {T_b} \propto m$
Or $\Delta {T_b} = \,{K_b}\,m$
Where, m= Molality of solution
To calculate the molar mass of the solute we have to use the above relation.
Complete step by step answer:
Step1: Molality is given by the formula,
$m = \dfrac{{No.\,of\,moles\,of\,solute}}{{Mass\,of\,solvent}}$
Number of moles in turn is given by the formula,
$Moles\,of\,solute = \dfrac{{Mass\,of\,solute}}{{Molar\,mass}}$
Substituting all the values in above relation we will get final formula,
$\Delta {T_b} = \dfrac{{1000 \times {K_b} \times {w_B}}}{{{w_A} \times {M_B}}}$ ……………..(i)
Where, $\Delta {T_b} =$ elevation in boiling point
${K_b} =$ molal boiling point elevation constant or ebullioscopic constant
${w_B} =$ weight of solute
${w_A} =$ weight of solvent
${M_B} =$ molar mass of solvent
Step2: In question we have given values :
${K_b} = 5.02Kkgmo{l^{ - 1}}$
${w_B} = 0.6126g$
${w_A} = 50.0g$
$\Delta {T_b} = 0.647K$
Step 3: From eq (i), molar mass of unknown solute is given as:
${M_B} = \dfrac{{1000 \times {K_b} \times {w_B}}}{{{w_A} \times \Delta {T_b}}}$
$= \dfrac{{1000 \times 5.03 \times 0.6216}}{{50.0 \times 0.647}}$
$= 96.65gmo{l^{ - 1}}$
Note: The elevation in boiling point is a colligative property which depends on the number of molecules but not on the nature of solute particles. The elevation in boiling point (\[\Delta {T_b}\]) is useful in determining the molar mass of the solute. It must be remembered that elevation in boiling point is a colligative property but boiling point is not a colligative property.
$\Delta {T_b} \propto m$
Or $\Delta {T_b} = \,{K_b}\,m$
Where, m= Molality of solution
To calculate the molar mass of the solute we have to use the above relation.
Complete step by step answer:
Step1: Molality is given by the formula,
$m = \dfrac{{No.\,of\,moles\,of\,solute}}{{Mass\,of\,solvent}}$
Number of moles in turn is given by the formula,
$Moles\,of\,solute = \dfrac{{Mass\,of\,solute}}{{Molar\,mass}}$
Substituting all the values in above relation we will get final formula,
$\Delta {T_b} = \dfrac{{1000 \times {K_b} \times {w_B}}}{{{w_A} \times {M_B}}}$ ……………..(i)
Where, $\Delta {T_b} =$ elevation in boiling point
${K_b} =$ molal boiling point elevation constant or ebullioscopic constant
${w_B} =$ weight of solute
${w_A} =$ weight of solvent
${M_B} =$ molar mass of solvent
Step2: In question we have given values :
${K_b} = 5.02Kkgmo{l^{ - 1}}$
${w_B} = 0.6126g$
${w_A} = 50.0g$
$\Delta {T_b} = 0.647K$
Step 3: From eq (i), molar mass of unknown solute is given as:
${M_B} = \dfrac{{1000 \times {K_b} \times {w_B}}}{{{w_A} \times \Delta {T_b}}}$
$= \dfrac{{1000 \times 5.03 \times 0.6216}}{{50.0 \times 0.647}}$
$= 96.65gmo{l^{ - 1}}$
Note: The elevation in boiling point is a colligative property which depends on the number of molecules but not on the nature of solute particles. The elevation in boiling point (\[\Delta {T_b}\]) is useful in determining the molar mass of the solute. It must be remembered that elevation in boiling point is a colligative property but boiling point is not a colligative property.
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