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Question: A solution of A (MM=20) and B (MM=10), [Mole fraction \({X_B}\) =0.6] having density 0.7 gm/ml th...

A solution of A (MM=20) and B (MM=10), [Mole fraction XB{X_B} =0.6] having density 0.7
gm/ml then molarity and molality of B in this solution will be ____ and ______ respectively.

Explanation

Solution

Here, we have to use the relation between molarity and mole fraction to calculate Molarity, that is, Molarity=xBd×1000xAMA+xBMB{\rm{Molarity}} = \dfrac{{{x_B}d \times 1000}}{{{x_A}{M_A} + {x_B}{M_B}}}. The another formula we have to use to calculate molality is Molality=XB×1000(1XB)MA{\rm{Molality}} = \dfrac{{{X_B} \times 1000}}{{\left( {1 - {X_B}} \right){M_A}}}, where, XA{X_A} is mole fraction of A, XB{X_B} is mole fraction
of B, MA{M_A} is molar mass of A, MB{M_B} is molar mass of B and d is density of the solution

Complete step by step answer:
Here, molar mass of A (MA)\left( {{M_A}} \right) is given as 20 and molar mass of B (MB)\left( {{M_B}} \right)
is 10.
The mole fraction of B (XB)\left( {{X_B}} \right) is given as 0.6. Now, we have to calculate the mole fraction
of A using the formula XA+XB=1{X_A} + {X_B} = 1.
Therefore, the mole fraction of A(XA)\left( {{X_A}} \right)
=1XB=10.6=0.41 - {X_B} = 1 - 0.6 = 0.4
Now, we have to put all the above values in the formula of molarity.
Molarity=xBd×1000xAMA+xBMB{\rm{Molarity}} = \dfrac{{{x_B}d \times 1000}}{{{x_A}{M_A} + {x_B}{M_B}}}
Molarity=0.6×0.7×10000.4×20+0.6×10\Rightarrow {\rm{Molarity}} = \dfrac{{0.6 \times 0.7 \times 1000}}{{0.4 \times 20 + 0.6 \times 10}}
Molarity=4208+6=30M\Rightarrow {\rm{Molarity}} = \dfrac{{420}}{{8 + 6}} = 30\,{\rm{M}}
Therefore, the molarity of the solution is 30 M.
Now, we have to calculate the molality of the solution. The relation between molality and mole fraction
is, Molality=XB×1000(1XB)MA{\rm{Molality}} = \dfrac{{{X_B} \times 1000}}{{\left( {1 - {X_B}} \right){M_A}}}
Now, we have to put the values of XB{X_B} = 0.6 and MA{M_A} =20 in the formula of molality.
Molality=XB×1000(1XB)MA{\rm{Molality}} = \dfrac{{{X_B} \times 1000}}{{\left( {1 - {X_B}} \right){M_A}}}
Molality=0.6×1000(10.6)20\Rightarrow {\rm{Molality}} = \dfrac{{0.6 \times 1000}}{{\left( {1 - 0.6} \right)20}}
Molality=75m\Rightarrow {\rm{Molality}} = 75\,{\rm{m}}
Therefore the molality of the solution is 75 m.

Additional Information:
Molarity changes with temperature as volume changes with respect to temperature. There are various
ways of expressing molarity, such as, decimolar, semimolar, millimolar etc. Molality is the better way of
expressing concentration than molarity because volume is not involved in molality. Therefore, no effect
of temperature is observed.

Note:
Always remember that both molarity and molarity are used to express concentration of solutions. Molarity is the moles of solute present in a solution of one litre and molality the moles of solute present in the solvent of one kg. The unit of molarity is M and the unit of molality is m.