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What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ?(Ag = 107.8, N = 14, O = 16, Na = 23, CI = 35.5)

  • Option 1)

    28 g

  • Option 2)

    3.5 g

  • Option 3)

    7 g

  • Option 4)

    14 g

 

Answers (1)

As we discussed in

Mass percent of solution -

Mass percent = ((mass of solute)/(mass of solution) ) X 100

 

- wherein

 it is mass of solute present in 100 gram solution.

 

 16.9% solution of AgNO3 means 16.9g AgNO3 in 100ml solution.

50ml of solution contains \frac{16.9}{2}\:=\:8.45g

Similarly, 100ml of other solution contains 5.8g NaCl

50ml of other solution contains 2.9g NaCl

Mol of AgNO3 = \frac{8.45}{169.8}\:=\:0.049

Mol of NaCl = \frac{2.9}{58.5}\:=\:0.049

AgNO3 + NaCl \rightarrow Ag0Cl + Na0NO3

0.049        0.049      0.049       0.049     

0               0              

Mass of AgCl = 0.049 x Molar mass of AgCl

                     = 0.049 x 143.5 = 7g


Option 1)

28 g

This option is incorrect.

Option 2)

3.5 g

This option is incorrect.

Option 3)

7 g

This option is correct.

Option 4)

14 g

This option is incorrect.

Posted by

Sabhrant Ambastha

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