Half litre of a solution contains 15% of alcohol. To change the alcohol concentration to 50%, find the quantity of alcohol to be mixed.

To find the quantity of alcohol to be mixed, we can set up a step-by-step algebraic equation based on the volumes of the components.

Step 1: Understand the initial state of the solution.
We are given that the initial volume of the solution is a half litre.
Since $1\text{litre}=1000\text{ml}$, the initial volume is:
$\text{Initial volume of solution}=500\text{ml}$

The initial concentration of alcohol in this solution is $15\%$. We can calculate the volume of alcohol currently present:
$\text{Volume of alcohol}=15\%\text{of}500\text{ml}$
$\text{Volume of alcohol}=\frac{15}{100}\times 500=75\text{ml}$

Step 2: Define the variable for the added alcohol.
Let $x$ be the quantity of pure alcohol (in ml) to be added to the mixture.

When we add $x$ ml of alcohol, both the volume of the alcohol and the total volume of the solution will increase by $x$ ml:
$\text{New volume of alcohol}=75+x$
$\text{New total volume of solution}=500+x$

Step 3: Set up the equation for the new concentration.
We want the final alcohol concentration to be $50\%$ (which is equivalent to $\frac{1}{2}$ in fraction form).
Therefore, the ratio of the new volume of alcohol to the new total volume of the solution is:
$\frac{75+x}{500+x}=\frac{50}{100}$
$\frac{75+x}{500+x}=\frac{1}{2}$

Step 4: Solve for $x$.
Cross-multiply to clear the fraction:
$2(75+x)=1(500+x)$
$150+2x=500+x$

Subtract $x$ from both sides:
$150+x=500$

Subtract $150$ from both sides:
$x=500-150$
$x=350\text{ml}$

Thus, $350\text{ml}$ of alcohol must be mixed into the solution to change its concentration to $50\%$.