Calculate the amount of water (in grams) that must be added to (a) 5.00 g of urea (NH2)2CO in the preparation of a 16.2 percent by mass solution, and (b) 26.2 g of MgCl2 in the preparation of a 1.5 percent by mass solution.

Respuesta :

Answer:

a. 25.8 g of water

b. 1720.4 g of water

Explanation:

A percent by mass, means the grams of solute in 100 g.  of solution. So, 16.2 g of urea are contained in 100 g of solution.

Then, solute mass + solvent mass = solution mass

16.2 g of urea + solvent mass = 100 g

solvent mass = 100 g - 16.2 g → 83.8 g

Now we can make the rule of three:

16.2 g of urea use 83.8 g of water

then, 5 g of urea would use (5 . 83.8) / 16.2 = 25.8 g of water

b. 1.5 % by mass, means 1.5 g of solute in 100 g of solution.

So water mass, for this solution will be 100 g - 1.5 g = 98.5 g

Now, we apply the rule of three:

1.5 g of solute use 98.5 of water

26.2 g of solute will use (26.2  . 98.5)/1.5 = 1720.4 g

The 5.00 g of urea in the preparation of a 16.2 percent by mass solution need 25.8 gram of water and 26.2 g of magnesium Chloride in the preparation of a 1.5 percent by mass solution needs 1720.4 g of water.

 

Percent by mass solution, means the grams of solute in 100 g. of solution.

(A) So, 16.2 g of urea are contained in 100 g of solution.

So, 83.8 water added to make 16.2 g solution.

Thus, 5 g of urea need

[tex]\bold {\dfrac { (5\times 83.8)} { 16.2} = 25.8 g}[/tex]

 

(B). 1.5 % by mass, means 1.5 g of solute in 100 g of solution.  

So, 98.5 g of water added to prepare 1.5 % solution.      

26.2 g of Magnesium Chloride will use

[tex]\bold {\dfrac {26.2 \times 98.5}{1.5} = 1720.4 g}[/tex]

Therefore, the 5.00 g of urea in the preparation of a 16.2 percent by mass solution need 25.8 gram of water and 26.2 g of magnesium Chloride in the preparation of a 1.5 percent by mass solution needs 1720.4 g of water.

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