How many kg of pure salt must be added to 8 kg of a 30% solution to obtain a 35% soltution? Round to the nearest tenth.
Let x be the number of kg added. the amount of NaCl initially present is
30% x 8 = 2.40 kg .
For the stronger mixture,
x + 2.4 = 0.35 (8.0 + x)= 2.8 + 0.35 x
0.65 x = 0.4
Solve for x
well consider the original solution.
IT has .3*8kg of salt in it, and .7*8kg water.
Newsalt+oldsalt=.35(8kg +newsalt)
.65Newsalt=.35*8 - .3*8=.4kg
solve for Newsalt.
To solve this problem, we can use the concept of the amount of salt present in the solution.
Let's break down the information provided:
- We have 8 kg of a 30% salt solution.
- We want to obtain a 35% salt solution.
To find the amount of pure salt in the original 8 kg solution, we can calculate:
Amount of salt = 8 kg * 30% = 2.4 kg
Now, let's assume we add x kg of pure salt to the solution.
The new total mass of the solution will be 8 kg + x kg, and the amount of salt will be 2.4 kg + x kg.
We want the new solution to be 35% salt, so we can set up the following equation:
(2.4 kg + x kg) / (8 kg + x kg) = 35%
Let's solve this equation algebraically:
(2.4 + x) / (8 + x) = 0.35 (converting 35% to decimal)
(2.4 + x) = 0.35 * (8 + x) (cross-multiplication)
Simplifying, we have:
2.4 + x = 0.35 * 8 + 0.35 * x
2.4 + x = 2.8 + 0.35x
0.65x = 0.4 (rearranging terms)
x = 0.4 / 0.65 ≈ 0.6154 kg (simplifying the equation)
Therefore, to obtain a 35% salt solution, we need to add approximately 0.6154 kg of pure salt to the 8 kg of the 30% salt solution.