Mathematics, 14.04.2020 19:02 jonnycampbell223

Solution A is an $80\%$ acid solution. Solution B is a $30\%$ acid solution. (a) Find the amount of Solution A (in mL) that must be added to $500$ mL of Solution B in order to produce a $70\%$ acid solution. (b) Find the amount of Solution A and Solution B (in mL) that can be combined in order to form a $100$ mL solution that is $50\%$ acid. (c) Does there exist a combination of Solution A and Solution B that is $90\%$ acid

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Answer from: paaoolav

(a) $2000\text{ ml}$ (b) $40\text{ ml and }60\text{ ml}$ (c)NO

Step-by-step explanation:

GIVEN: Solution A is an $\$80\%$ acid solution. Solution B is a $\$30\%$ acid solution.

TO FIND: (a) Find the amount of Solution A (in mL) that must be added to $500 \text{ml}$ of Solution B in order to produce a $70\%$ acid solution. (b) Find the amount of Solution A and Solution B (in mL) that can be combined in order to form a $100 \text{ml}$ solution that is $50\%$ acid. (c) Does there exist a combination of Solution A and Solution B that is $90\%$ acid.

SOLUTION:

(a)

Let total amount of solution A added be

Total amount of mixture $=500+x\text{ ml}$

As resulting solution is $70\%$ acid solution.

Amount of acid in final solution is sum of acid in both solutions

$\frac{70}{100}(500+x)=\frac{80}{100}x+\frac{30}{100}\times500$

35000+70x=80x+15000

$x=2000\text{ ml}$

(b)

Let amount of Solution A be $x\text{ ml}$

Amount of solution B $=100-x\text{ ml}$

As resulting solution is $50\%$ acidic solution

Amount of acid in final solution is sum of acid in both solutions

$\frac{50}{100}\times100=\frac{80}{100}x+\frac{30}{100}(100-x)$

5000=80x+3000-30x

50x=2000

$x=40\text{ ml}$

Amount of solution B $=100-40=60\text{ ml}$

(c)

No as concentration of final solution can not be greater than one with higher concentration.

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Solution A is an $80\%$ acid solution. Solution B is a $30\%$ acid solution. (a) Find the amount of...