If 0.1 M solution of NaCl is isotonic with 1.1 w % urea solution, the degree of ionisation of NaCl is (molar masses of urea and NaCl are 60 and 58.5 g `mol ^(-1)` respectively )

Options

A. 2
B. 0.83
C. 1
D. 1.83

Answer

B

Explanation

Given,
Concentration of NaCl (C) = 0.1 M
Mass of urea `(w_B)` = 1.1g
Molar mass of urea `(M_B)` = 60 g `mol^(-1)`
Molar mass of NaCl `(M_A)` = 58.5 g `mol^(-1)`
Volume of solution (V) = 100 mL
`:'` Solutions of NaCl and urea are isotonic.
`:." "pi(NaCl)=pi("urea")`
`ixxCRT(NaCl)=CRT(urea)`
where,
i = van't Hoff factor, C = concentration
R = gas constant, T = temperature
`:." " ixxC(NaCL)=C("urea")`
`ixx0.1/(1L)=w_B/m_Bxx1000/V`
`i=1.1/60xx1000/100xx1/0.1=1.83`
also, for ionisation `(i)=1-alpha+nalpha`
where, `alpha` = degree of dissociation
n = number of particles = 2( for NaCl)
Thus,
`i=1-alpha+2alpha`
`1.83=1+alpha`
or, `alpha=1.83-1.0`
Degree of dissociation `(alpha)` =0.83
Hence, option (b) is the correct answer.