Question

10. How much alcohol must be added to 500 mL of a 60% solution to make its strength 75%?
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Answer 1

Answer:

Solution = 400 ml

Strength = 32 %

Let x ml of alcohol be added

We need to calculate the quantity of alcohol

Using formula for quantity

Quantity of alcohol in 400 ml solution Q =\dfrac{15}{100}\times400Q=

100

15

×400

Q=60\ mlQ=60 ml

We need to calculate the value of x

Using formula for strength

\dfrac{60+x}{400+x}=\dfrac{32}{100}

400+x

60+x

=

100

32

100\times(60+x)=32(400+x)100×(60+x)=32(400+x)

6000+100x=12800+32x6000+100x=12800+32x

100x-32x=12800-6000100x−32x=12800−6000

68x=680068x=6800

x=\dfrac{6800}{68}x=

68

6800

x=100\ mlx=100 ml

Hence, 100 ml pure alcohol must be added .

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Answer 2

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