Question

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

100 g of 0.500% (w/w) NaI
250 g of 0.500% (w/w) NaBr
500 g of 1.25% (w/w) C6H12O6 (glucose)
750 g of 2.00% (w/w) H2SO4

Answer 1

Step-by-step 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}}* 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\%=(x)/(100 g)* 100`

`x=(0.500* 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\%=(x)/(250 g)* 100`

`x=(0.500* 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\%=(x)/(500 g)* 100`

`x=(1.25* 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\%=(x)/(750 g)* 100`

`x=(2.00* 750 g)/(100)=15.0 g`

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