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Question: Ammonia in \(0.224gm\) of a compound, \(Zn{(N{H_3})_x}C{l_2}\) is neutralised by \(30.7ml\) of \(0.2...

Ammonia in 0.224gm0.224gm of a compound, Zn(NH3)xCl2Zn{(N{H_3})_x}C{l_2} is neutralised by 30.7ml30.7ml of 0.20M0.20M HClHCl . The value of xx in the formula is:
A.4
B.5
C.6
D.8

Explanation

Solution

A chemical substance composed of atoms of more than one element which is held together by chemical bonds is known as chemical compound. Mainly these exist in two categories, organic compounds (carbon and its derivatives) and inorganic compounds.

Complete step by step answer:
Given:
Weight of compound = 0.224gm0.224gm
Concentration of HClHCl = 0.20M0.20M
Volume of HClHCl required to neutralize the compound = 30.7ml30.7ml
Molecular weight of Zn(NH3)xCl2Zn{(N{H_3})_x}C{l_2} =65.30+17x+35.5×2  =136.30+17x  gm/mol65.30 + 17x + 35.5 \times 2\; = 136.30 + 17x\;gm/mol
136.30+17x  gm136.30 + 17x\;gm of compound contains xx mole of ammonia.
Therefore,
0.224gm0.224gm of compound contains x17x+136.30×0.224\dfrac{x}{{17x + 136.30}} \times 0.224 moles.
And we know that xx equivalent of ammonia = xx mole of ammonia
So,
Eq. of HClHCl =MV1000== \dfrac{{MV}}{{1000}} =Eq. of NH3N{H_3}
0.224x17x+136.30=30.7×0.21000\dfrac{{0.224x}}{{17x + 136.30}} = \dfrac{{30.7 \times 0.2}}{{1000}}
x=6.756x = 6.75 \approx 6
We will take the integral value of xx because the molecule can’t be fractional.
So the obtained integer value of xx is 66 .

Hence option (C) is correct.

Note:
Equivalent is basically the amount of substance that reacts with an arbitrary amount of another substance in a given chemical reaction. It is also defined as the mass of a substance which displaces one gram of hydrogen or eight gram of oxygen from a compound.