A solution of 17.1 w% of sucrose ( molar mass `= 342 " g mol"^(-1)`) is isotonic with a `x ` w% solution of oxalic acid ( molar mass `= 90g " mol"^(-1)`). Assume the degree of dissociation of oxalic acid as 0.01. What is `x` ?

Options

A. 9
B. 0.45
C. 4.41
D. `0.90`

Answer

C

Explanation

Given, Mass of sucrose `(w_(1))=17.1 g`
Molar mass of sucrose `(M_(1))=342g "mol"^(-1)`
Mass of oxalic acid `(w_(2))=xg`
Molar mass of oxalic acid `(M_(2))=90g "mol"^(-1)`
Degree of dissociation `(alpha)` of oxalic acid `=0.01`
step I To find the value of van.t Hoff factor (i)
`i=(1-alpha)+n alpha`
where, `n=` number of parts in which one molecule of oxalic acid dissociates `(=3)`
`i=(1-0.01)+(3xx0.01)`
`i=0.99+0.03`
`i=1.02`
Step II To find the value of `(x)`
For isotonic solutions :
`pi` ( sucrose) `= pi` ( oxalic acid) [ `:. pi = ` CRT]
or, Molarity ( sucrose) `=i` Molarity ( oxalic acid )
`(w_(1))/(M_(1))` ( sucrose) `= i xx ((x)/(M_(2)))` ( oxalic acid )
`x= (w_(1))/(M_(1))xx (M_(2))/(i)= (17.1xx90)/(1.02xx342)`
`x=(1539)/(348.84)=4.41g`
Hence, value of `x=4.41 g`
Hence, option (c) is correct.