Answer:
The concentration of the most dilute solution is 0.016M.
Explanation:
First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:
![[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M](https://tex.z-dn.net/?f=%5BNa_%7B2%7DSO_%7B4%7D%5D%3D%5Cfrac%7Bmoles%28Na_%7B2%7DSO_%7B4%7D%29%7D%7Bliters%28solution%29%7D%20%3D%5Cfrac%7Bmass%28%28Na_%7B2%7DSO_%7B4%7D%29%29%7D%7Bmolarmass%28moles%28Na_%7B2%7DSO_%7B4%7D%29%20%5Ctimes%200.100L%29%7D%20%3D%5Cfrac%7B2.5316g%7D%7B142g%2Fmol%5Ctimes%200.100L%20%7D%20%3D0.178M)
<u>First dilution</u>
We can use the dilution rule:
C₁ x V₁ = C₂ x V₂
where
Ci are the concentrations
Vi are the volumes
1 and 2 refer to initial and final state, respectively.
In the first dilution,
C₁ = 0.178 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
<u>Second dilution</u>
C₁ = 0.053 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
Answer:
1 kg=1000g so 2000 g=2kg and 1 liter of water is equal to 1 kg .as a result 2 kg=2l and if consider 1l as 1000 ml we get 2000 ml
the final answer is 2000 g of water equal to 2000 ml
Answer:
1) 18.91 mL
2) 2.35 %
Explanation:
1. The volume of base that was required is equal to the <em>difference between the reading at the end of the titration and the reading at the beginning</em>:
- V = 20.95 mL - 2.04 mL = 18.91 mL
2. The mass percent can be written as:
- Mass Percent = Mass solute / Total mass * 100%
For this problem:
We<u> input the data and calculate the mass percent</u>:
- % mass = 2.28/96.92 * 100% = 2.35 %
