How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?

Answer 1

55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles.

20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles.

molarity of final diluted solution=22.9/20.5=1.117M

since the no. of moles of KI present in initial and final solution are same.

let.V(in ml) be the final volume of diluted solution.

222.75/V=1.117

V=199.41 ml

final volume =199.41 ml

Answer 2

you use the equation M1V1=M2V2

multiply (.355L)(1.25M)/(4.30M)= .103L of KI

1. (.355L)(1.25M)=(4.30M)(?L)

2. Divide each side by 4.30 M

3. (.355L) (1.25M)/4.30= .103 KI

Answer 3

the answer is 319 ml or rounding correctly it is 320 ml

Answer 4

The volume is 268,3 mL.

Answer 5

The solution contain 12,47 g KI.

Answer 6

0.0177grams