Answer:
Option C
Explanation:
Given,
mass of sucrose (w1) =1.71 g
Molar mass of sucrose (M1)=342 g mol-1
Mass of oxalic acid (w2) =xg
Molar mass of oxalic acid (M2) = 90 g mol-1
Degree of dissociation (α) of oxalic acid =0.01
Step I To find the value of van't Hoff factor (i)
i=(1−α)+nα
where, n= number of parts in which one molecule of oxalic acid dissociates (=3)
i=(1-0.01)+(3 x 0.01)
i= 0.99+0.03
i=1.02
Step II To find the value of (x)
For isotonic solutions:
π (sucrose) = π ( oxalic acid) [ ∵]
or, Molarity (sucrose) = i Molarity (oxalic acid)
\frac{w_{1}}{M_{1}} (sucrose) = i \times ( \frac{x}{ M_{2}}) (oxalic acid )
x= \frac{w_{1}}{M_{1}}\times\frac{M_{2}}{i}=\frac{17.1\times90}{1.02\times342}
x= \frac{1539}{348.84}=4.41 g
hence, value of x=4.41 g
Hence, option (c) is correct.