2. A solid chloride sample weighing 0.2853 g required 43.75 mL of 005273 M AgNO,

Question

2. A solid chloride sample weighing 0.2853 g required 43.75 mL of 005273 M AgNO, to reach the Ag,Cro, end point. a. How many moles Cl-ion were present in the sample? (Use Eqs. 2 and 3.) moles Cl- b. How many grams Ch ion were present? (Use Eq.4.) g Cl- c. What was the mass percent Cr ion in the sample? (Use Eq, 5.) How would the following errors affect the mass percent Cl obtained in Question 2c? Give your reason- ing in each case. 3. a. The student read the molarity of AgNO, as 0.05723 M instead of 0.05273 M b. The student was past the end point of the titration when he took the final buret reading.

Explanation / Answer

answer:

a)

moles of chloride = moles of AgNO3(at end point)

Moles of AgNO3 = Molarity ×Volume(in litres)

= 0.05273×43.75/1000=0.0023 = moles of chloride

b)  mass of chloride = moles * molar mass

= 0.0023 * 35.5 = 0.082g ( molar mass of chloride = 35.5 g)

c)  

% of chloride = (mass of chloride / total mass of sample)* 100

=(0.082 / 0.2853) * 100 = 28.7%

3)

a) when student will read given value (o.5723) of AgNO3, then moles of Cl will change becauew at end point moles of AgNO3 = moles of Cl

so, mole of Cl increase means mass of Cl also increase so mass pewrcetage of Cl will increase.

b)

at the end point, if the student done fastly then it is very difficult to identify end point.

according to M1V1 = M2V2 formula we get different concetration of Cl-

b)

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.