It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and 88 wt.% water. two solutions are available, the first contains 5 wt.% ethanol, and the second contains 25 wt.% ethanol. how much of each solution are mixed to prepare the desired solution?

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

Explanation:

Let the mass of the first solution be x and second solution be y.

Amount solution required = 1250 kg

x + y = 1250 kg....[1]

Percentage of ethanol in required solution = 12% of 1250 kg

Percentage of ethanol in solution-1 = 5% of x

Percentage of ethanol in required solution = 25% of y

5% of x +  25% of y =12% of 1250 kg

x + 5y = 3000 kg...[2]

Solving [1] and [2] we :

x = 437.5 kg   , y =  812.5 kg

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

if the temperature drops and becomes cool

then the atmosphere can be freezing as the water vapor in the air is at maximum (100%) and can be severely cold

also can be snowing lightly as the moisture in the air will start to solidify

and btw this is not chemistry but