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Solids in Liquids

5.85 g of N...

Question

5.85 g of $N a C l$ was treated with concentrated $H_{2} S O_{4}$ and the gas evolved was passed into a solution of silver nitrate. The white precipitate obtained was filtered, dried and weighed. Assuming complete reaction, how many grams of precipitate was obtained?
[Atomic mass of $A g = 108$ u, $N a = 23$ u and $C l = 35.5$ u]

10.8g

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14.45g

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35.5g

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3.65g

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Solution

The correct option is B 14.45g
Molecular weight of $N a c l = 58.5 g$ $\begin{matrix} Moles in 5.85 g & = \frac{5.85}{58.5} = 0.1 moles \end{matrix}$
Nacl $+$ conc. $H_{2} {S O}_{4} \to {N a H S O}_{4} + H C l$
0.1 moles
According to the above reaction,
0.1 moles of Nacl will form
0.1 moles of HCl gas.
when $H C l$ gas pass-through
silver nitrate solution,
Following reaction will
occur.
AgNO3 + HCL $\to$ Agcl + HNO $_{3}$
In the above reaction $0.1 m o l$
AgCl will be formed as
white pet.
Mass of 0.1 moles agcl
$\begin{matrix} = 0.1 \times (108 + 35.5) = 0.1 \times 143.5 = 14.35 g \end{matrix}$
Hence, 143.5 grams of white
ppt. will be formed.

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