Explanation
Molecular weight of $$C{{O}_{2}}=12+16+16=44$$
Molecular weight of $${{O}_{2}}=16+16=32$$
Molecular weight of $$C{{H}_{4}}=12+4=16$$
Molecular weight of $$S{{O}_{2}}=32+16+16=64$$
So;
$$S{{O}_{2}}$$has maximum Molecular weight.
Hence $$S{{O}_{2}}$$ exert maximum amount of partial pressure.
Given,
volume - $$0.820\ l$$
mass - $$1.00\ g$$
pressure - $$1.00\ atm$$
temperature - $${3^{\rm O}}C$$
Using ideal gas equation - $$PV = nRT$$
$$n$$ - number of mole = $$\dfrac{m}{M}$$
$$m$$ - given mass
$$M$$ - molar mass
$$PV = \dfrac{m}{M}RT$$
here, find the molar mass of the compound
$$M = \dfrac{{mRT}}{{PV}}$$
Firstly, change the temperature in Kelvin.
$${3^{\rm O}}C + 273 = 276K$$
Now, put the value in the formula.
$$M = \dfrac{{1 \times 0.0820 \times 276}}{{1 \times 0.820}}$$
= $$27.6\ g/mole$$
write each compound molar mass according to the given value and match it with this value.
(a) $$B{H_3}$$ - $$10.8 + 1 \times 3 = 13.8$$
(b) $${B_4}{H_{10}}$$ - $$4 \times 10.8 + 1 \times 10 = 53.2$$
(c) $${B_2}{H_6}$$ - $$2 \times 10.8 + 6 \times 1 = 27.6$$
(d) $${B_3}{H_{12}}$$ - $$3 \times 10.8 + 12 \times 1 = 44.4$$
So, option (c) is the correct answer.
As no of moles of $$O_2=16/32=0.5$$ moles
no of moles of $$N_2=28/28=1$$ moles
no of moles of $$CH_4=8/16=0.5$$ moles
Total pressure=740mm
Partial pressure=total pressure$$\times$$ mole fraction of $$N_2$$
Mole fraction of $$N_2$$ = No of moles of $$N_2$$ /total no of moles
$$=1/0.5+0.5+1=1/2$$
then partial pressure $$=1/2\times740$$
$$=370\ mm$$
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