A solution of two components containing `n_1` moles of `1^(st)` component and `n_2` moles of the `2^("nd")` component is prepared . `M_1` and `M_2` are the molecular weights of component 1 and 2 respectively . If d is the density of the solution in g` mL^(-1)` , `C_2` is the molarity and `x_2` is the mole fraction of the `2^("nd")` component , then `C_2` can be expressed as :

Options

A. `C_2=(1000x_2)/(M_1+x_2(M_2-M_1))`
B. `C_2=(d x_2)/(M_2+x_2(M_2-M_1))`
C. `C_2=(1000dx_2)/(M_1+x_2(M_2-M_1))`
D. `C_2=(dx_1)/(M_2+x_2(M_2-M_1))`

Answer

C

Explanation

To express `C_2` in terms of mole fraction `x_2`
`{:(,1^(st)"component",2^(nd) "component"),("Mol",n_1,n_2),(M.w.,M_1,M_2),("Mass",n_1M_1,n_2M_2):}`
Mass of solution `= n_1 M_1 + n_2M_2`
Mole fraction `x_2=(n_2)/(n_1+n_2)`
`n_1=(n_2(1-x_2))/x_2`
Volume of solution `=n_1M_1+n_2M_2`
`=(n_2M_1(1-x_2))/(x_2)+n_2M_2`
`=n_2/x_2[M_2x_2-x_2M_1+M_1]`
Volume of solution
`=(n_2[M_2x_2-x_2M_1+M_1])/(1000dx_2)`litre
`C_2=(1000n_2dx_2)/(n_2[M_2x_2-x_2M_1=M_1])`
`C_2=(1000n_2dx_2)/(M_1+x_2(M_2-M_1))`