JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 18)
A solution of two components containing
n1 moles of the 1st component and n2 moles of
the 2nd component is prepared. M1 and M2 are
the molecular weights of component 1 and 2
respectively. If d is the density of the solution
in g mL–1, C2 is the molarity and x2 is the mole
fraction of the 2nd component, then C2 can be
expressed as :
$${C_2} = {{1000{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
$${C_2} = {{1000d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
$${C_2} = {{d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
$${C_2} = {{d{x_1}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
Explanation
| 1st component | 2nd component | |
|---|---|---|
| Mole | n1 | n2 |
| Molecular Weight |
M1 | M2 |
| Mass | n1M1 | n2M2 |
Mass of solution = n1M1 + n2M2
density of the solution = d
C2 = Molarity of 2nd Component
x2 = Mole-fraction of 2nd Component
x2 = $${{{n_2}} \over {{n_1} + {n_2}}}$$
$$ \Rightarrow $$ 1 - x2 = $${{{n_1}} \over {{n_1} + {n_2}}}$$
Molarity of 2nd Component,
C2 =
n2
Volume of solution(ml)
$$ \times $$ 1000
= $${{{n_2}} \over {{{Mass\,of\,Solution(ml)} \over {Density\,of\,Solution(g/ml)}}}} \times 1000$$
= $${{d \times {n_2} \times 1000} \over {{n_1}{M_1} + {n_2}{M_2}}}$$
= $${{d \times {{{n_2}} \over {{n_1} + {n_2}}} \times 1000} \over {{{{n_1}} \over {{n_1} + {n_2}}}{M_1} + {{{n_2}} \over {{n_1} + {n_2}}}{M_2}}}$$ (Dividing componendo and dividendo by n1 + n2)
= $${{d \times {x_2} \times 1000} \over {\left[ {\left( {1 - {x_2}} \right){M_1} + {x_2}{M_2}} \right]}}$$
= $${{1000d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
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