 # In Zeisel’s method for the determination of methoxy groups, a sample of 2.68gm of a compound (A) gave 14.08gm of $AgI$. If the molecular weight of compound (A) is 134, the number of $\left( { - OC{H_3}} \right)$ group(s) in the compound (A) is-(A) 1(B) 2(C) 3(D) 4 Verified
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Hint: We will first understand what is meant by Zeisel’s method. Then we will undergo the reaction which the compound goes through while using this method. We will then simply find out the resultant weight of the $AgI$ precipitate formed.

As we know that the Zeisel’s method is used to count the number of methoxy groups in that compound. We use the agents that will only react with the methoxy groups in that compound to count its number.
In Zeisel’s method, we simply heat the given compound with hydriodic acid which in turn results in iodoalkanes. After doing so, we pass this iodoalkane through an ethanolic solution in the presence of silver nitrite $\left( {AgN{O_3}} \right)$. This results in a precipitate of $AgI$. We measure the weight of this $AgI$ precipitate which gives us the number of methoxy groups present. In other words, the number of moles of $AgI$ will be equal to the number of $\left( { - OC{H_3}} \right)$ groups.
Now, in the question, 2.68gm of compound A is given. This compound A gives us 14.08 gm of $AgI$.
Now we will calculate the number of methoxy groups in one mole.
Now, we already know that 2.68gm of A $\to$ 14.08gm of $AgI$
So, 134gm of A $\to$ $\dfrac{{14.08}}{{2.68}} \times 134$
$\Rightarrow 704gm$ of $AgI$ precipitate
Thus, the number of molecules in $AgI$-
Molecular weight of $AgI$ is 235.
$\Rightarrow \dfrac{{704}}{{235}} \\ \\ \Rightarrow 3moles \\$
This means that the number of methoxy groups are 3.
Hence, it is clear that option C is the correct option.

Note: Methanol and dimethyl ether are the simplest methoxy compounds. Anisole and vanillin are other methoxy ethers. Most alkoxides contain methoxy groups, e.g. orthosilicate tetramethyl and methoxide titanium. These compounds frequently are known as methoxides. 