theotherguyxxx
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A chemistry teacher needs to make 10 L of 36% sulphuric acid solution. The acid solutions available are 30% sulphuric acid and 50% sulphuric acid, by volume. How many litres of each solution must be mixed to make the 36% solution?

Respuesta :

Answer:7 L and 3 L

Step-by-step explanation:

Given

Solution [tex]1[/tex]  has [tex]30\%[/tex] sulphuric acid by volume

Solution [tex]2\ \text{has}\ 50\%[/tex] sulphuric acid by volume

final solution is [tex]10\ L[/tex] of [tex]36\%[/tex] sulphuric acid

Suppose there is [tex]V_1\ L[/tex] and [tex]V_2\ L[/tex] of sulphuric acid of solution [tex]1[/tex] and [tex]2[/tex] respectively

So, quantity of acid before and after remains same

[tex]0.3\times V_1+0.5\times V_2=0.36\times 10[/tex]

[tex]3V_1+5V_2=36\quad \ldots(i)[/tex]

and total mixture before and after remains same i.e.

[tex]V_1+V_2=10\quad \ldots(ii)[/tex]

Solving [tex](i)[/tex]  and  [tex](ii)[/tex] we get

So we get [tex]V_1=7\ L[/tex] and [tex]V_2=3\ L[/tex]

So, 7 liters from solution 1 and 3 liters from solution 2

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