Question

What quantities of 90% V/v and 50% v/v
alcohol are mixed to make 400 ml of 60%
V/v alcohol?​

Answer 1

Answer:

your answer is in the attachment

Explanation:

hope it's help you XD

Answer 2

Given,

The concentration of the alcohol with a higher concentration = 90%$\frac{v}{v}$

The concentration of the alcohol with a lower concentration = 50%$\frac{v}{v}$

The concentration of the mixture = 60%$\frac{v}{v}$

The volume of mixture = 400ml

To Find,

The volume of 90% and 50% alcohol to be mixed.

Solution,

The alligation method can be used for the calculation of quantities of the two different solutions to be mixed to produce a solution of the required quantity.

According to the alligation method, the proportion of the solution with a higher concentration in the mixture= $Required concentration- lower concentration$

Similarly, the proportion of the solution with a lower concentration in the mixture = $Higher concentration-Required concentration$

∴ The proportion of 90% alcohol in the mixture = $Required conc. - lower conc.$

⇒$60-50 =10 parts$

The proportion of 50% alcohol in the mixture = $Higher conc. - Required conc.$

⇒$90-60 =30 parts$

The volume of 90% alcohol in 400ml 0f 60%$\frac{v}{v}$ alcohol = $\frac{Parts of given conc.}{total parts} (total volume)$

⇒ $\frac{10}{40} (400ml)$

⇒ $100 ml$

The volume of 50% alcohol in 400ml 0f 60%$\frac{v}{v}$ alcohol =$\frac{Parts of given conc.}{total parts} (total volume)$

⇒ $\frac{30}{40} (400ml)$

⇒$300ml$

Henceforth, the volumes of 90% and 50% alcohol to be mixed are 100ml and 300ml respectively.