Respuesta :
Explanation:
(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.
[tex]w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]
1) 100 g of 0.500% (w/w) NaI
Mass of solution = 100 g
Mass of solute = x
Required w/w % of solution = 0.500%
[tex]0.500\%=\frac{x}{100 g}\times 100[/tex]
[tex]x=\frac{0.500\times 100 g}{100}=0.500 g[/tex]
0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.
2) 250 g of 0.500% (w/w) NaBr
Mass of solution = 250 g
Mass of solute = x
Required w/w % of solution = 0.500%
[tex]0.500\%=\frac{x}{250 g}\times 100[/tex]
[tex]x=\frac{0.500\times 250 g}{100}=1.25 g[/tex]
1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr
3) 500 g of 1.25% (w/w) glucose
Mass of solution = 500 g
Mass of solute = x
Required w/w % of solution = 1.25%
[tex]1.25\%=\frac{x}{500 g}\times 100[/tex]
[tex]x=\frac{1.25\times 500 g}{100}=6.25 g[/tex]
6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)
4) 750 g of 2.00% (w/w) sulfuric acid.
Mass of solution = 750 g
Mass of solute = x
Required w/w % of solution = 2.00%
[tex]2.00\%=\frac{x}{750 g}\times 100[/tex]
[tex]x=\frac{2.00\times 750 g}{100}=15.0 g[/tex]
15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.
