InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
What is the mass of precipitate formed when 50 mL of 16.9% solution of silver nitrate is mixed with 50 mL of 5.8% NaCl solution? (Ag = 107.8 g, N = 14, O = 16, Na = 23, Cl = 35.5) |
|
Answer» Solution :50 mL of `16.9%AgNO_(3)` solution contaion `AgNO_(3)` `=(16.9)/(100)xx50=8.45g` 50 mL of `5.8%` NACL solution contains NaCl `=(5.8)/(100)xx50=2.9g` `{:(""AgNO_(3),+,NaCl,rarr,AgCl + NaNO_(3)),(107.8+14+48,+,23+35.5,,108+35.5),(=169.8g,,=58.5g,,=143.5g):}` `8.45" g of "AgNO_(3)` reacts with `NaCl=(58.5)/(169.8)xx8.45g` `=2.9g` Thus, the amounts present react completely `therefore"AgCl FORMED"=(143.5)/(169.8)xx8.45g~=7g` |
|