Question

What is the mass of precipitate formed when 50 ml of 16.9 solution of agno3?

Answer 1

Well, w/v≡Mass of soluteVolume of solution , we should get over 7⋅g of silver ... And thus mass of silver nitrate=16.9%×50⋅mL=8.45⋅g .

Answer 2

Answer:

Moles of AgNo3= 50x 16.9/100x169.8

= 0.05mole

Moles of NaCl = 50x5.8/100x58.5

=0.05mole

AgNo3 + NaCl ----> AgCl + NaNo3

mass of AgCl = mole x molar mass

=0.05 x 143.5

= 7.16 g

Hope it help uuu..