Question

Q18 What volume of 2M HCI should be mixed with 8M HCl to get two litres of 4M HCI? O 1.0L 0 0.50L 1.33L O 0.25L​

Answer 1

Explanation:

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Answer 2

Given - M1 = 2M

M2 = 8M

M3 = 4M

V3 = 2 litres

Find - V1 and V2 required to get V3 which is 2 litres.

Solution - Using the formula M1V1 + M2V2 = M3V3.

Let us assume the V1 to be x. Therefore, V2 will be (2 - x).

Keeping the values in formula

2*x + 8*(2 - x) = 4*2

2x + 16 - 8x = 8

16 - 8 = 8x - 2x

8 = 6x

x = 8/6

x = 1.33

V2 = 2 - x

V2 = 2 - 1.33

V2 = 0.67

Therefore, 1.33 litre of 2M HCl is required and 0.67 litre of 8M HCl is required.