what is the mass of precipitate formed when 50mL of 16.9% solution of AgNO3 is mixed with 50mL of 5.8% NaCl solution?(Ag=
Hello Aspirant,
AgNO3
Given
Ag = 107.8g, N = 14g, O = 16g
Molar mass = [107.8 + 14 + (16×3)] g = 169.8 g
16.9% solution implies 16.9 g of AgNO3 in 100 mL of solution.
So, 1 mL solution comprises (16.9/100) g of AgNO3.
So, 50 mL solution contains [(16.9/100)× 50 ] g of AgNO3.
Hence, 50 mL solution contains 8.45g of AgNO3.
No.of moles = (given mass/molar mass)
= [8.45g /(169.8 g/mol) ]= 0.049 moles = 0.05 moles (approx)
NaCl
Molar mass = (23 + 35.5 )g = 58.5 g.
5.8% solution implies 5.8g of NaCl in 100 mL of solution.
So, 1 mL solution comprises (5.8/100) g of NaCl.
So, 50 mL solution contains [(5.8/100)× 50 ] g of NaCl.
Hence, 50 mL solution contains 2.9g of NaCl.
No.of moles = (given mass/molar mass)
= (2.9/58.5) = 0.049 = 0.05 (approx)
AgNO3(aq) + NaCl(aq) AgCl↓ (s) +NaNO3(aq)
ppt.
The molar mass of AgCl = (107.8+35.5) g = 143.3g.
Mass of AgCl ppt. formed = Mole × Molar mass =0.05 moles ×143.3 g =7.165 g
Therefore, 7.165 g of AgCl will be formed.
Regards.
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