Concentration - salt solution

We added 300 g of water to 400 g of salt solution, reducing the solution's concentration by 5%. How much water did the original solution contain, and what was its concentration?

Final Answer:

w1 =  353.33 g
p1 =  11.6667 %

Step-by-step explanation:

m1​=400 g Δw=300 g  m2​=m1​+Δw=400+300=700 g  p1​ =  m1​s​ ⋅  100 p2​ =  m2​s​ ⋅  100 p2​ = p1​ −5  s/m2​ ⋅  100 = s/m1​⋅ 100−5  s/m2​=s/m1​−5/100 s/700=s/400−5/100  3s=140  s=3140​=46.66666667  s=3140​≐46.666667  m1​ = w1​+s w1​=m1​−s=400−46.6667=353.33 g
p1​=m1​s​⋅ 100=40046.6667​⋅ 100=335​≐11.6667%   Verifying Solution:  p2​=m2​s​⋅ 100=70046.6667​⋅ 100=320​≐6.6667 % Δp=p1​−p2​=11.6667−6.6667=5 %

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algebrabasic operations and conceptsUnits of physical quantitiesthemes, topicsGrade of the word problem

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