
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
457.2k+ 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.
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