Question

Calculate the number of grams of solute needed to make each of the following solutions:

Answer 1

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

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%

$0.500\%=\frac{x}{100 g}\times 100$

$x=\frac{0.500\times 100 g}{100}=0.500 g$

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%

$0.500\%=\frac{x}{250 g}\times 100$

$x=\frac{0.500\times 250 g}{100}=1.25 g$

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%

$1.25\%=\frac{x}{500 g}\times 100$

$x=\frac{1.25\times 500 g}{100}=6.25 g$

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%

$2.00\%=\frac{x}{750 g}\times 100$

$x=\frac{2.00\times 750 g}{100}=15.0 g$

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.