for an experiment a student requires 200milliliters of a solution that is 11% Na

Question

for an experiment a student requires 200milliliters of a solution that is 11% NaCl. the storeroom has only solutions that are 15% NaCl and 5% NaCl. how many milliliters of each available solution should be mixed to get 200 milliliters of 11% NaCl? for an experiment a student requires 200milliliters of a solution that is 11% NaCl. the storeroom has only solutions that are 15% NaCl and 5% NaCl. how many milliliters of each available solution should be mixed to get 200 milliliters of 11% NaCl? for an experiment a student requires 200milliliters of a solution that is 11% NaCl. the storeroom has only solutions that are 15% NaCl and 5% NaCl. how many milliliters of each available solution should be mixed to get 200 milliliters of 11% NaCl?

Explanation / Answer

Let x milliliters of 15% NaCl and y milliliters of 5% NaCl be mixed.

So in 15% of x ml would be 15x/100 NaCl

And 5% of y ml would be 5y/100 NaCl

So from the given conditions, we have x+y=200-------(1)

11% of 200 ml would be 11*200/100=22ml of NaCl

So 15x/100+5y/100=22

3x/20+y/20=22

3x+y=440---(2)

Solving equations 1 and 2, we get:

x=120ml and y=80ml

We mix 120ml of 15% solution and 80ml of 5% solution.

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