Question

A sample of 1.79 mg of a compound of molar mass $90g\,mo{l^{ - 1}}$. When treated with $C{H_3}Mgl$ releases 1.34 ml of a gas at STP. The number of active hydrogens in the molecule is:
A.1
B.2
C.3
D.4

Answer 1

Hint: The equation formed will be \[R{(OH)_n} + nC{H_3}Mgl \to nC{H_4} + R{(OMgl)_n}\], and according to equation, 1 mole of \[C{H_4}\] gas is equivalent to 1mole of active H atoms. We will find the moles of the compound since its mass and molar mass are given. We will then find the moles of \[C{H_4}\], and for finding the number for moles of active Hydrogen (H) atoms per mole of compound, we will calculate \[\dfrac{{nC{H_4}}}{{{n_{compound}}}}\].

Complete step by step answer:
Since hydrogen is released, that means the compound may be alcohol or acid
1.79 mg of a compound releases 1.34 ml of hydrogen gas
Let the reaction be \[R{(OH)_n} + nC{H_3}Mgl \to nC{H_4} + R{(OMgl)_n}\]
According to the above equation, 1 mole of \[C{H_4}\] gas is equivalent to 1 mole of active H atoms. Thus, the moles of \[C{H_4}\] gas are equal to the moles of active H atoms
 \[{n_{compound}} = \] Mass of compound / molar mass of compound \[ = \dfrac{{0.46g}}{{92gmo{l^{ - 1}}}} = 0.005\]
$ \Rightarrow $ \[nC{H_4} = \dfrac{{336mole}}{{22400mLmo{l^{ - 1}}}} = 0.015\]
Thus, the number for moles of active Hydrogen (H) atoms per mole of compound is =
 \[\dfrac{{nC{H_4}}}{{{n_{compound}}}}\]
\[ = \dfrac{{0.015}}{{0.005}} = 3\]
Thus, there are active Hydrogen (H) atoms per molecule of compound.

Therefore, the correct answer is option (C).

Note: We can use an alternate method to solve this question:
We know that, hydrogen gas is released, that means the compound may be alcohol or acid
1.79 mg of a compound releases 1.34 ml of hydrogen gas
At STP, 22400 ml is the volume occupied by 1 mole of \[{H_2}\]
Amount of gas produced \[ = {\text{ }}\dfrac{{1.34ml}}{{22400MLmo{l^{ - 1}}}} = 6 \times {10^{ - 5}}mol\]
Amount of compound used $ = \dfrac{1}{\pi }(6 \times {10^{ - 5}}mol)$
This is equivalent to \[ = \left( {\dfrac{{1.79 \times {{10}^{ - 3}}g}}{{90gmo{l^{ - 1}}}}} \right) = 2 \times {10^{ - 5}}mol\]
Hence, \[n = \dfrac{{6 \times {{10}^{ - 5}}mol}}{{2 \times {{10}^{ - 5}}mol}} = 3\]
 \[n = 3\]
 $ \Rightarrow $ Active Hydrogen (H) atoms per molecule of compound is 3.