Question

What is the mass of precipitate formed when 50 ml of 8.5 % solution of AgNO3 is mixed with 100 ml of 1.865 % potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g

Answer 1

^_^ MOLE CONCEPT

I think You Provided % w/v not %w/w ^_^

Final Answer : a) 3.59 g

Steps:
1) Molar Mass of AgNO3 ,
= 108 + 14 + 3*16 = 170g
Molar Mass of KCl = 39 + 35.5 = 74.55 g
Molar Mass of AgCl = 108 + 35.5 =143.55g
Since,
%x w/v denotes x g of solute dissolved in 100 mL of solution /water.

Accordingly,
Mass of AgNO3 = 8.5 * 50/100 = 4.25 g
Mass of KCl = 1.865 g


2) We have Reaction ^.^,

AgNO3 + KCl -----> AgCl. + KNO3

no. of moles of AgCl = no. of moles reactant reacted.

no. of moles of AgNO3 = 4.25 / 170= 0.025
no. of moles of KCl =1.865 / 74.55= 0.025 (approx)

3) Hence,
No. of moles of AgCl (PPT) = 0.025
Mass of AgCl = no. of Moles * Molar Mass
= 0.025* 143.55
= 3.59 g (approx)


Hence, Mass of PPT obtained is 3.59 g

Answer 2

Answer:

a)3.59 g

Mass of AgNO3 is equal to number of moles multiple molar mass