Answer:
The student should use 156.7 g of solution
Explanation:
31.9 %w solution of acetic acid in ethanol means 100 g of solution contains 31.9 g of acetic acid.
So, 450. g of solution contains [tex](\frac{31.9\times 450.}{100})g[/tex] of acetic acid or 143.55 g of acetic acid
Alternatively, 143.55 g of acetic acid is present in 450. g of solution
So, 50.00 g of acetic acid is present in [tex](\frac{450.\times 50.00}{143.55})g[/tex] solution or 156.7 g of solution
Hence the student should use 156.7 g of solution