Question

200 g of 25% sulphuric acid solution was added to 300 g of 40% sulphuric acid solution Find the concentration of the acid in the
mixture
A 14%
B. 24%
C. 44%
D. 34%​

Answer 1

Answer:

Step-by-step explanation:

(200×25)+(300×40)=17000

200+300=500

17000÷500=34%

Answer 2

Answer: D) 34%

Step-by-step explanation:

Solution 1: 200 g have 25% of Sulphuric acid

Therefore amount of sulphuric acid

$=25\% \text { of  }200\\\\= \dfrac{25}{100} \times 200 =50g$

Solution 2 : 300 g have 40 % sulphuric acid

therefore amount of sulphuric acid

$=40\% \text { of} 300 \\\\= \dfrac{40}{100} \times 300=120g$

Total solution = solution 1+ solution 2 = 200 + 300 = 500

suphuric acid = 50 + 120 = 170 g

Concentration of acid in mixture

$\dfrac{acid}{solution} \times 100= \dfrac{170}{500} \times 100 = 34\%$

Hence the concentration of acid in mixture is 34%