An aqueous solution of \[4\% \] non-volatile solute exerts a pressure of $1.004bar$ at the normal boiling point of the solvent. What is the molar mass of the solute?
$\left( A \right)41.35gmo{l^{ - 1}}$
$\left( B \right)60.21gmo{l^{ - 1}}$
$\left( C \right)84.42gmo{l^{ - 1}}$
$\left( D \right)56.91gmo{l^{ - 1}}$
Answer
585.3k+ views
Hint: A solution is made up of solvent and solute. non-volatile solute means which does not evaporate . It has high boiling points and less vapour pressure . we can also say that substances which have boiling point ${100^ \circ }C$ or more are termed as volatile rest all are non-volatile .
Complete answer:
Vapour pressure of water at normal boiling point given is $1.013bar$
Vapour pressure given for solution in normal boiling point is $1.004bar$
Given that solute \[4\% \] means $4g$
Mass of solvent will be $100 - 4 = 96g$
Molar mass of water is $18gmo{l^{ - 1}}$
Now we have to apply Raoult’s law
$ \Rightarrow $ $\dfrac{{{p_1}^0 - {p_1}}}{{{p_1}^0}} = \dfrac{{{w_2} \times {M_1}}}{{{M_2} \times {w_1}}}$
Which states that vapour pressure of a solution of non-volatile solute should be equal to the vapour pressure of the pure solvent at that temperature which is multiplied with mole fraction.
Putting all the values
$ \Rightarrow $ $\dfrac{{1.013 - 1.004}}{{1.013}} = \dfrac{{2 \times 18}}{{{M_2} \times 96}}$
${M_2} = 84.42gmo{l^{ - 1}}$
Hence the molar mass of the solute is $84.42gmo{l^{ - 1}}$
So the correct answer is Option B.
Additional information:
There are four types of colligative properties which are: vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure. Colligative is a Latin word which means bounds together.
Colligative properties are those which depend upon the number of solute particles not on the nature of the solution. Basically it does not affect the types of solute present in the solvent only the number of solutes are actually present that matters.
Note:
Remember Raoul’s law only works for lower concentration . Also the interactions of solute and solvent must be identical. As the intermolecular forces vary a lot between the molecules so only by keeping the concentration of solute low we can minimise the intermolecular forces.
Complete answer:
Vapour pressure of water at normal boiling point given is $1.013bar$
Vapour pressure given for solution in normal boiling point is $1.004bar$
Given that solute \[4\% \] means $4g$
Mass of solvent will be $100 - 4 = 96g$
Molar mass of water is $18gmo{l^{ - 1}}$
Now we have to apply Raoult’s law
$ \Rightarrow $ $\dfrac{{{p_1}^0 - {p_1}}}{{{p_1}^0}} = \dfrac{{{w_2} \times {M_1}}}{{{M_2} \times {w_1}}}$
Which states that vapour pressure of a solution of non-volatile solute should be equal to the vapour pressure of the pure solvent at that temperature which is multiplied with mole fraction.
Putting all the values
$ \Rightarrow $ $\dfrac{{1.013 - 1.004}}{{1.013}} = \dfrac{{2 \times 18}}{{{M_2} \times 96}}$
${M_2} = 84.42gmo{l^{ - 1}}$
Hence the molar mass of the solute is $84.42gmo{l^{ - 1}}$
So the correct answer is Option B.
Additional information:
There are four types of colligative properties which are: vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure. Colligative is a Latin word which means bounds together.
Colligative properties are those which depend upon the number of solute particles not on the nature of the solution. Basically it does not affect the types of solute present in the solvent only the number of solutes are actually present that matters.
Note:
Remember Raoul’s law only works for lower concentration . Also the interactions of solute and solvent must be identical. As the intermolecular forces vary a lot between the molecules so only by keeping the concentration of solute low we can minimise the intermolecular forces.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

