Question

What is the mass of the precipitate formed when
50 ml of 16.9% (W/V) solution of AgNO, is mixed
with 50 mL of 5.8% (w/v) Naal solution?
(Ag = 1078, N = 14,0 = 16, N = 22.1 35.5)





solution plz​

Answer 1

Answer:

AgNO

3

+NaCl→AgCl+NaNO

3

ForAgNO

3

,

V=50ml

V

W

%=16.9%

50

W

=

100

16.9

W

AgNO

3

=8.45g

n

AgNO

3

=

M

AgNO

3

W

AgNO

3

=0.05

ForNaCl,

V=50ml

V

W

%=5.8%

W

NaCl

=2.9g

n

NaCl

=

M

NaCl

W

NaCl

=0.05

AccordingtoreactionAgCl(ppt)=0.05moles×MolarmassofAgCl=0.05×142.8=7.14g≈7g

Answer 2

Solution :-

AgNO3

• Volume of solution (V)= 50 ml

• % W /V = 16.9 %

• Molar mass of AgNO3 = 170 g( approx.)

we know that ,

⟹ % W/V = mass of the solute / volume of solution x 100

⟹ 16.9= w x100/ 50

⟹ w = 8.45 g

Now that we have the mass of AgNO3 let us find the moles ,

⟹ n = w / M

here ,

• n = moles

• w = mass of solute

• M = molar mass of solute

⟹ n = 8.45 / 170

⟹ n = 0.05

NaCl

• Volume of solution (V')= 50 ml

• % W '/V '= 5.8%

• Molar mass of NaCl = 58 g( approx.)

we know that ,

⟹ % W/V = mass of the solute / volume of solution x 100

⟹ 5.9 = w' x100/ 50

⟹ w '= 2.95 g

Now that we have the mass of NaCl let us find the moles ,

⟹ n '= w' / M '

here ,

• n = moles

• w = mass of solute

• M = molar mass of solute

⟹ n' = 2.95 / 58

⟹ n '= 0.05

Reaction :-

AgNO3 + NaCl ➝ AgCl (ppt) + NaNO3

we know that ,

⟹ moles of reactant / stoichiometry = moles of product / stoichiometry

⟹ moles of AgNO3  / 1 = moles of AgCl  / 1

⟹ moles of AgNO3   = moles of AgCl

⟹ moles of AgCl (n") = 0.05

• Molar mass of AgCl = 143 g

we know that ,

⟹ n" = w " / M "

On rearranging ,

⟹ w " = n" x M"

⟹ w " = 0.05 x 143

⟹ w" = 7.15 g

The mass of precipitate formed is 7.15 g