contestada

to compound 50% solution in which proportion must 95% sol and 30% sol be mixed to make
500 ml

Respuesta :

Answer:

4/9

Step-by-step explanation:

Let the amount of 95% sol = a ml and

the amount of 30% sol = b ml.

So, the total amount of the compound=a+b=500 ml

As the compound is 50% solution, so

95% of a + 30% of b= 50% of (a+b)

(95/100)a+(30/100)b=(50/100)(a+b) [from equation (i)]

95a + 30b = 50(a+b)

95(a/b)+30 = 50(a+b)/b [dividing both sides by b]

95(a/b)+30 = 50(a/b+1)

95(a/b)+30 = 50(a/b)+50

95(a/b)-50(a/b)=50-30

45(a/b)=20

a/b=20/45

a/b=4/9

Therefore, a:b=4:9

As a+b= 500 ml, so the amount of 95% sol = [tex]\frac {500}{13}\times 4=\frac{2000}{13}=153.85[/tex] ml

and the amount of 95% sol = [tex]\frac {500}{13}\times 9=\frac{4500}{13}=346.15[/tex] ml

Hence, the proportion of 95% sol and 30% sol to make a compound of 50% sol must be 4/9.

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