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Troyanec [42]
4 years ago
9

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
Mathematics
1 answer:
Irina-Kira [14]4 years ago
7 0

Answer:

(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 x

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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