Answer :
Final answer:
To produce a solution that is 35.0 wt% SrCl2 with 36.0 grams of SrCl2, one would need to add 66.9 grams of water. This is because the total weight of the solution is the weight of SrCl2 plus the weight of the water needed.
Explanation:
In order to determine how much water must be added to 36.0 g of SrCl2 to produce a solution that is 35.0 wt% SrCl2, we must understand that the 35% by weight represents that 35 g of SrCl2 is in 100 g of the solution. Thus, to find the total weight of the water and the SrCl2, we should set up the equation 36g (weight of SrCl2) / X g (total weight of solution) = 35%, which would give that X = 102.9 g. The total solution weight is the weight of the SrCl2 plus the weight of the water needed, so to find the weight of the water we subtract the weight of SrCl2 (36 g) from the total weight of solution (102.9 g), which gives us 66.9 g as the amount of water to be added to the SrCl2.
Learn more about Solution Preparation here:
https://brainly.com/question/37304944
#SPJ11
To solve this problem we will use the following equation:
w = (m of solute) / (m of solution)
w - percentage
It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.
w = mass CaCl2/(mass of water + mass of CaCl2)
mass of water = x
0.35 = 36 / (x + 36)
0.35 × (x + 36) = 36
0.35x + 12.6 = 36
0.35x = 23.4
x = 66.86 g of water is necessary
w = (m of solute) / (m of solution)
w - percentage
It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.
w = mass CaCl2/(mass of water + mass of CaCl2)
mass of water = x
0.35 = 36 / (x + 36)
0.35 × (x + 36) = 36
0.35x + 12.6 = 36
0.35x = 23.4
x = 66.86 g of water is necessary
