Answer:
0.21 M. (2 sig. fig.)
Explanation:
The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.
How many moles of NaOH in the original solution?
,
where
is the number of moles of the solute in the solution.
is the concentration of the solution. 
for the initial solution.
is the volume of the solution. For the initial solution, 
for the initial solution.
.
What's the concentration of the diluted solution?
.
- is the number of solute in the solution. Diluting the solution does not influence the value of .
for the diluted solution. - Volume of the diluted solution:
.
Concentration of the diluted solution:
.
The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
Therefore, we have;
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.
Answer:
Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.
However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.
The molar concentration of the original HF solution : 0.342 M
Further explanation
Given
31.2 ml of 0.200 M NaOH
18.2 ml of HF
Required
The molar concentration of HF
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)
Titrant = NaOH(1)
Titrate = HF(2)
Input the value :
