Answer:
120 g of NaCl in 300 g H20 at 90 C
Explanation:
At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40
we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g
Answer:
237.8L of water would need to be added.
Explanation:
The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).
Then figure out what information you have and what is being found. In this case:
M1 = 54.7 M
V1 = 1092 mL = 1.092 L
M2 = 0.25 M
V2 = unknown
Then solve the equation for whatever you are trying to find.
M1V1=M2V2
V2=M1V1/M2
Now you need to plug everything in.
V2=(54.7M*1.091L)/0.25M
V2=238.93L
That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.
238.93L - 1.091L = 237.8L
Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.
I hope this helps. Let me know if anything is unclear.
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