Some Basic Concepts of Chemistry - Result Question 42
####42. What is the mass of precipitate formed when 50 $mL$ of $16.9 %$ solution of $AgNO_3$ is mixed with 50 $mL$ of $5.8 % NaCl$ solution? [2015 RS] $(Ag=107.8, N=14, O=16, Na=23, Cl=35.5)$
(a) $28 g$
(b) $3.5 g$
(c) $7 g$
(d) $14 g$
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Solution:
- (c) $50 mL$ of $16.9 %$ solution of $AgNO_3$
$ \begin{aligned} & (\frac{16.9}{100} \times 50)=8.45 g of^{2 g NO_3} \\ & n _{\text{mole }}=\frac{8.45 g}{(107.8+14+16 \times 3) g / mol} \\ & =(\frac{8.45 g}{169.8 g / mol})=0.0497 moles \end{aligned} $
$50 mL$ of $5.8 %$ solution of $NaCl$ contain
$NaCl=(\frac{5.8}{100} \times 50)=2.9 g$ $n _{NaCl}=\frac{2.9 g}{(23+35.5) g / mol}=0.0495 moles$
$AgNO_3+NaCl \to AgCl \downarrow+Na^{+}+NO_3^{-}$
$ 1 \text{ mole } \quad 1 \text{ mole } \quad 1 \text{ mole } $
$ \therefore \quad 0.049 \text{ mole } \quad 0.049 \text{ mole } \quad 0.049 \text{ mole of } AgCl $
$n=\frac{w}{M} \to w=(n _{\text{AgCl }}) \times$ Molecular Mass
$ =(0.049) \times(107.8+35.5)=7.02 g $
