Question

100 ml of 1M HCL 200 ml of 2M HCL 300ml of 3M HCL are mixed with enough water to get 1M solutions the volume of water to be added is​

Answer 1

Hey mates your answer is here

Answer:The volume of the solution is 14 mL.

Explanation

Number of moles on 1 M HCL solution in 100 mL

Number of moles on 2 M HCL solution in 200 mL

Number of moles on 3 M HCL solution in 300 mL

On mixing all the solution with water to get 1 M solution HCl.

Total number of moles in resulting solution = 0.1 +0.4+ 0.9 = 1.4 moles

Volume of the resulting 1 M HCl solution = V

The volume of the solution is 14 mL.

May be it's helpful for you

please mark me brainliest ✌️✌️

Answer 2

In the above question the amount of water added is 800ml

Given,

100ml of 1M OF HCL

200ml of 2M OF HCL

300ml of 3M OF HCL

Assume,

1M = M1

2M = M2

3M = M3

100ml = V1

200ml = V2

300ml = V3

1M = M

$mv = \frac{ m_{1} v_{1} + m_{2} v_{2} + m_{3}v_{3}}{m}$

$= \frac{1 \times 100 + 2 \times 200 + 3 \times 300}{1} ml$

=100+400+900 ml

= 1400ml

$v = v_{1} + v_{2} + v_{3} + v_{ h_{2}o}$

Therefore

$v_{h _{2}o} = v - (v_{1} + v_{2} + v_{3})$

=(1400-(100+200+300)) ml

= 800ml