What mass of water should be added to 22.0 g of KCl to make a 5.50 % by mass solution?

We have that for the Question "" it can be said that the  mass of water should be added to 22.0 g of KCl to make a 5.50 % by mass solution is

W=378.5gram

From the question we are told

What mass of water should be added to 22.0 g of KCl to make a 5.50 % by mass solution?

Generally the equation for the water needed  is mathematically given as

Therefore

The  mass of water should be added to 22.0 g of KCl to make a 5.50 % by mass solution is

W=5.5 +378

W=378.5grams

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383.5 grams is the mass of water should be added to 22.0 g of KCl to make a 5.50 % by mass solution.

Explanation:

Given that:

Solution to be prepared is 5.5 %

mass of KCl is 22 gram

mass of water required =?

5.5 % solution means that

5.5 grams in 100 ml or gram of water to make solution.

let the water required is 5.5 + x, so the equation will be

5.5 = x 100

121+ 5.5x = 2200

5.5 x = 2200-121

     x =

    x = 378 grams

The mass of water added to KCl to make it 5.5% percent solution is

5.5 +378

= 383.5 grams.

to make the solution of 5.5 percent of KCl weighing 22 grams.

indicate valleys, hills, steepness and gentleness of slopes.

explanation:

contour lines on a map indicate elevation, or depth. numeric notations on contour lines refer to the level of elevation, or height about sea level.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2