<em>C = 0,75 mol/dm³</em>
<em>V = 500mL = 500cm³ = 0,5dm³</em>
C = n/V
n = 0,75×0,5dm³
<u>n = 0,375 moles</u>
<em>M NaCl: 23+35,5 = 58,5g</em>
1 mole ---------- 58,5g
0,375 ----------- X
X = 0,375×58,5
<u>X = 21,9375g NaCl</u>
:)
Answer:
The answer is 0.0698 M
Explanation:
The concentration was prepared by a serial dilution method.
The formula for the preparation I M1V1 = M2V2
M1= the concentration of the stock solution = 0.171 M
V1= volume of the stock solution taken = 200 mL
M2 = the concentration produced
V2 = the volume of the solution produced = 940 mL
Substitute these values in the formula
0.171 × 200 = 490 × M2
34.2 = 490 × M2
Make M2 the subject of the formula
M2 = 34.2/490
M2 = 0.069795
M2 = 0.0698 M ( 3 s.f)
The concentration of the Chemist's working solution to 3 significant figures is 0.0698M
This allows the water cycle to take hold. Transferring water to places all across the globe, adding minerals and purifying other water In the process.