- Column-I amp-Column-II- i- 0-5 mole of SO-2-g- amp-p- Occupies 11-2 L at 1 atm and 273 K- ii- 1 g of H-2-g- amp-q- Weighs 24 g- iii- 0-5 mole of O-3-g- amp-r- Total atoms-1-5 - N-A- iv- 1 g molecule of O-2-g- amp-s- Weighs 32 g- -i- - p nbsp- ii- p-q nbsp- iii- - p nbsp- iv- - si- - p- r- s nbsp- ii- p nbsp- iii- - p- q nbsp- iv- - ri- - p- r nbsp- ii- p nbsp- iii- - p- r nbsp- iv- - si- - p- r- s nbsp- ii- p nbsp- iii- - p- q- r nbsp- iv- - s

The correct option is D i) p, r, s nbsp; ii) p nbsp; iii) p, q, r nbsp; iv) sAnd the answer is : i) 0.5 moles of SO_2(g) nbsp; 1 mole of SO_2(g) ...

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Byju's Answer

Standard XI

Mathematics

Number of Elements in a Cartesian Product

[ ...

Question

$\begin{matrix} Column-I & Column-II i) 0.5 mole of S O_{2} (g) & p) Occupies 11.2 L at 1 atm and 273 K ii) 1 g of H_{2} (g) & q) Weighs 24 g iii) 0.5 mole of O_{3} (g) & r) Total atoms = 1.5 \times N_{A} iv) 1 g molecule of O_{2} (g) & s) Weighs 32 g \end{matrix}$

i) - p, r, s ii)- p iii) - p, q, r iv) - s

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i) - p ii)- p,q iii) - p iv) - s

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i) - p, r, s ii)- p iii) - p, q iv) - r

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i) - p, r ii)- p iii) - p, r iv) - s

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Solution

The correct option is A i) - p, r, s ii)- p iii) - p, q, r iv) - s
And the answer is :
i) 0.5 moles of $S O_{2} (g)$
1 mole of $S O_{2} (g)$ occupies $22.4 L$ at STP
0.5 moles of $S O_{2} (g)$ occupies $11.2 L$ at STP
$Number of atoms = Moles \times N_{A} \times atomicity$
The number of atoms in 0.5 moles of $S O_{2} (g) = 0.5 \times N_{A} \times 3 = 1.5 N_{A}$
1 mole of $S O_{2}$ weighs $= 32 + 2 \times 16 = 64 g$
0.5 moles of $S O_{2}$ weighs $32 g$ .
Thus, p, r, s are correct match for i).

ii) 1 g of $H_{2} (g)$
$Number of moles = \frac{Given mass}{Molar mass}; Given mass = 1 g$ and the $molar mass = 2 g/mol$
Number of moles $= \frac{1}{2} = 0.5 mol$
1 mole of $H_{2} (g)$ occupies $22.4 L$ at STP
0.5 moles of $H_{2} (g)$ occupies $11.2 L$ at STP
Thus, p is the correct match for ii).

iii) 0.5 moles of $O_{3} (g)$
1 mole of $O_{3} (g)$ occupies $22.4 L$ at STP
0.5 mole of $O_{3} (g)$ occupies $11.2 L$ at STP
Weight of 1 mole of $O_{3} (g) = 16 \times 3 = 48 g$
Weight of 0.5 moles of $O_{3} (g) = 24 g$
$Number of atoms = Moles \times N_{A} \times atomicity$
The number of atoms in 0.5 moles of $O_{3} (g) = 0.5 \times N_{A} \times 3 = 1.5 N_{A}$
Thus, p, q, r are correct match for iii).

iv) 1 g molecule of $O_{2}$ means 1 mole of $O_{2}$ = $2 \times 16 = 32 g$
Thus, s is a correct match for iv).

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Column-IColumn-IIi) 0.5 mole of SO2(g)p) Occupies 11.2 L at 1 atm and 273 Kii) 1 g of H2(g)q) Weighs 24 giii) 0.5 mole of O3(g)r) Total atoms=1.5×NAiv) 1 g molecule of O2(g)s) Weighs 32 g

$\begin{matrix} Column-I & Column-II i) 0.5 mole of S O_{2} (g) & p) Occupies 11.2 L at 1 atm and 273 K ii) 1 g of H_{2} (g) & q) Weighs 24 g iii) 0.5 mole of O_{3} (g) & r) Total atoms = 1.5 \times N_{A} iv) 1 g molecule of O_{2} (g) & s) Weighs 32 g \end{matrix}$