Match List-I and List-II List-I List-II [Volume at S.T.P]a14g of N2i4.48 lit b1.2044×1023 molecules N2ii22.4 litc18 gm of H2O iii18 mLd9.033×1023 atoms of Helliumiv11.2 lit v33.6 lit(Given molar volume of a gas at STP = 22.4lit)Correct matching is

a) Volume of 14g of N2=1428×22.4=11.2litb) Volume of 1.2044×1023 molecules N2=1.2044×10236.022×1023×22.4=4.48litc) Volume of 1 mole of H2O at S.T.P= 18 mld) Volume of 9.033×1023 atoms of He =9.033×10236.022×1023×22.4=33.6lit

Match List-I and List-II List-I List-II [Volume at S.T.P]a14g of N2i4.48 lit b1.2044×1023 molecules N2ii22.4 litc18 gm of H2O iii18 mLd9.033×1023 atoms of Helliumiv11.2 lit v33.6 lit(Given molar volume of a gas at STP = 22.4lit)Correct matching is

a) Volume of 14g of N2=1428×22.4=11.2litb) Volume of 1.2044×1023 molecules N2=1.2044×10236.022×1023×22.4=4.48litc) Volume of 1 mole of H2O at S.T.P= 18 mld) Volume of 9.033×1023 atoms of He =9.033×10236.022×1023×22.4=33.6lit

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Match List-I and List-II

List-I List-II
[Volume at S.T.P]
a $14 g of N_{2}$ i 4.48 lit
b $1.2044 \times 10^{23}$ molecules $N_{2}$ ii 22.4 lit
c 18 gm of $H_{2} O$ iii 18 mL
d $9.033 \times 10^{23}$ atoms of Hellium iv 11.2 lit
v 33.6 lit
(Given molar volume of a gas at STP = 22.4lit)
Correct matching is

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a-iv,b-i, c- iii, d-v

a-iv,b-ii, c- v, d-iii

a-ii ,b-i, c- iii, d-v

a-v,b-i, c- ii, d-iv

answer is A.

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Detailed Solution

a) Volume of 14g of $N_{2} = \frac{14}{28} \times 22.4 = 11.2 lit$
b) Volume of $1.2044 \times 10^{23}$ molecules $N_{2} = \frac{1.2044 \times 10^{23}}{6.022 \times 10^{23}} \times 22.4 = 4.48 l i t$
c) Volume of 1 mole of $H_{2} O$ at S.T.P= 18 ml
d) Volume of $9.033 \times 10^{23}$ atoms of He $= \frac{9.033 \times 10^{23}}{6.022 \times 10^{23}} \times 22.4 = 33.6 l i t$

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	List-I		List-II [Volume at S.T.P]
a	$14 g of N_{2}$	i	4.48 lit
b	$1.2044 \times 10^{23}$ molecules $N_{2}$	ii	22.4 lit
c	18 gm of $H_{2} O$	iii	18 mL
d	$9.033 \times 10^{23}$ atoms of Hellium	iv	11.2 lit
		v	33.6 lit

Match List-I and List-II List-I List-II [Volume at S.T.P]a14g of N2i4.48 lit b1.2044×1023 molecules N2ii22.4 litc18 gm of H2O iii18 mLd9.033×1023 atoms of Helliumiv11.2 lit v33.6 lit(Given molar volume of a gas at STP = 22.4lit)Correct matching is

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