Question

What is the mass of the precipitate formed when $50mL$ of $16.9\%$ solution of ${\text{AgNO}}_3$ is mixed with $50mL$ of $5.8\%$ w/v $\text{NaCl}$ solution?

$[Ag=107.8,N=14,O=16,Na=23,Cl=35.5]$

• A. $7g$
• B. $14g$
• C. $28g$
• D. $3.5g$

Answer 1

The correct option is A $7g$

The molar masses of${\text{AgNO}}_3,\text{NaCl}$ and $\text{AgCl}$ are $169.8$ g/mol, $58.5$ g/mol and $143.3$ g/mol respectively.

The number of moles of ${\text{AgNO}}_3$ present is

$\frac{50\times 16.9}{100\times 169.8}=0.0497$

The number of moles of $\text{NaCl}$ present is

$\frac{50\times 5.8}{100\times 58.5}=0.0496$

Thus, both ${\text{AgNO}}_3$ and $\text{NaCl}$ are present in the stoichiometric quantity.

The number of moles of $\text{AgCl}$ formed is $0.0496$.

The mass of $\text{AgCl}$ formed is $0.0496\times 143.3=7$ g.

Hence, the correct option is $A$