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Question: A sample of 1.79 mg of a compound of molar mass \(90g\,mo{l^{ - 1}}\). When treated with \(C{H_3}Mgl...

A sample of 1.79 mg of a compound of molar mass 90g mol−190g\,mo{l^{ - 1}}. When treated with CH3MglC{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

Explanation

Solution

The equation formed will be R(OH)n+nCH3Mgl→nCH4+R(OMgl)nR{(OH)_n} + nC{H_3}Mgl \to nC{H_4} + R{(OMgl)_n}, and according to equation, 1 mole of CH4C{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 CH4C{H_4}, and for finding the number for moles of active Hydrogen (H) atoms per mole of compound, we will calculate nCH4ncompound\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+nCH3Mgl→nCH4+R(OMgl)nR{(OH)_n} + nC{H_3}Mgl \to nC{H_4} + R{(OMgl)_n}
According to the above equation, 1 mole of CH4C{H_4} gas is equivalent to 1 mole of active H atoms. Thus, the moles of CH4C{H_4} gas are equal to the moles of active H atoms
ncompound={n_{compound}} = Mass of compound / molar mass of compound =0.46g92gmol−1=0.005 = \dfrac{{0.46g}}{{92gmo{l^{ - 1}}}} = 0.005
⇒\Rightarrow nCH4=336mole22400mLmol−1=0.015nC{H_4} = \dfrac{{336mole}}{{22400mLmo{l^{ - 1}}}} = 0.015
Thus, the number for moles of active Hydrogen (H) atoms per mole of compound is =
nCH4ncompound\dfrac{{nC{H_4}}}{{{n_{compound}}}}
=0.0150.005=3= \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 H2{H_2}
Amount of gas produced = 1.34ml22400MLmol−1=6×10−5mol = {\text{ }}\dfrac{{1.34ml}}{{22400MLmo{l^{ - 1}}}} = 6 \times {10^{ - 5}}mol
Amount of compound used =1π(6×10−5mol) = \dfrac{1}{\pi }(6 \times {10^{ - 5}}mol)
This is equivalent to =(1.79×10−3g90gmol−1)=2×10−5mol = \left( {\dfrac{{1.79 \times {{10}^{ - 3}}g}}{{90gmo{l^{ - 1}}}}} \right) = 2 \times {10^{ - 5}}mol
Hence, n=6×10−5mol2×10−5mol=3n = \dfrac{{6 \times {{10}^{ - 5}}mol}}{{2 \times {{10}^{ - 5}}mol}} = 3
n=3n = 3
⇒\Rightarrow Active Hydrogen (H) atoms per molecule of compound is 3.