(a) A gas cylinder of capacity of $20 d m^{3}$ is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass fo the hydrogen is 2 g, hence the relative molecular mass of the gas is :
(i) 5 (ii) 10 (iii) 15 (iv) 20
(b) (i) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens beause fo the heat evolved while calcium carbide reacts with and acetylene gas is formed. If $200 c m^{3}$ of acetylene is formed from a certain mass of calcium carbidem, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.
$2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)$
(ii) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogenj by mass. Find the molecular formula of the ocmpound if its relative molecular mass is 37. $[N = 14, H = 1] .$
(c) (i) A gas cylinder contains $24 \times 10^{24}$ moleules of nitrogen gas. If avogadro's number is $6 \times 10^{23}$ and the relative atomic mass of nitrogen is 14, calculate:
(1) Mass of nitrogen gas in th ecylinder
(2) Volume of nitrogen at STP in $d m^{3}$
(ii) Commercial sodium hydroxide wieghing 30 g has some sodium chloride in it. The mixture on dissolving in water and subsequent treatment with excesss silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage fo sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is
$N a C l + A g N O_{3} \to A g C l + N a N O_{3}$
[Relative molecular mass of $N a C l = 58; A g C l = 143$ ]
(iii) A certain Gas 'X' occupies a volume of $100 c m^{3}$
at S.T.P. and weighs 0.5 g. Find its relative moleular mass.

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Solution

$left parenthesis a right parenthesis M a s s space o f space g a s space X equals 10 g M a s s space o f space h y d r o g e n space g a s space equals 2 R e l a t i v e space v a p o u r space d e n s i t y space equals fraction numerator M a s s space o f space v o l u m e space o f space g a s space X space u n d e r space s i m i l a r space c i n d i t i o n s over denominator M a s s space o f space v o l u m e space o f space h y d r o g e n space g a s space u n d e r space s i m i l a r space c o n d i t i o n end fraction equals 10 over 2 equals 5 R e l a t i v e space m o l e c u l a r space m a s s space o f space t h e space g a s space equals 2 cross times r e l a t i v e space v a p o u r space d e n s i t y equals 2 cross times 5 equals 10 left parenthesis b right parenthesis left parenthesis i right parenthesis 2 C subscript 2 H subscript 2 left parenthesis g right parenthesis plus 5 O subscript 2 left parenthesis g right parenthesis rightwards arrow 4 C O left parenthesis g right parenthesis plus 2 H subscript 2 O left parenthesis g right parenthesis A c c o r d i n g space t o space G a y minus L u s s a c apostrophe s space l a w 2 space v o l u m e space o f space a c e t y l e n e space r e q u i r e s space 5 space v o l u m e space o f space o x y g e n space t o space b u r n space i t space t h e r e f o r e space 1 space v o l u m e space o f space a c e t y l e n e space r e q u i r e s space 2.5 space v o l u m e space o f space o x y g e n space t o space b u r n space i t 200 space c m cubed space r e q u i r e s space 2.5 cross times 200 equals 500 c m cubed space o f space o x y g e n 2 space v o l u m e space o f space a c e t y l e n e space o n space c o m b u s t i o n space g i v e s space 4 C O subscript 2 t h e r e f o r e space 1 space v o l u m e space o f space a c e t y l e n e space o n space c o m b u s t i o n space g i v e s space 2 space C O subscript 2 200 space c c space o f space a c e t y l e n e space o n space c o m b u s t i o n space w i l l space g i v e space 200 cross times 2 equals 400 space c c space o f space C O subscript 2 left parenthesis ii right parenthesis H y d r o g e n space equals 12.5 percent sign t h e r e f o r e space N i t r o g e n space equals 100 minus 12.5 equals 87.5 percent sign$

Element % weight Atomic weight Atomic ratio simple ratio

N 87.5 14 87.5/14=6.25 1

H 12.5 1 12.5/1=12.5 2

The Empirical formula of the compound is NH2
Empirical formula weight =14+2=16
Relative molecular mass=37
$N equals fraction numerator R e l a t i v e space m o l e c u l a r space m a s s over denominator E m p i r i c a l space w e i g h t end fraction equals 37 over 16 equals 2.3 almost equal to 2 M o l e c u l a r space f o r m u l a space equals n cross times e m p i r i c a l space f o r m u l a space equals 2 cross times N H subscript 2 equals N subscript 2 H subscript 4 left parenthesis c right parenthesis left parenthesis i right parenthesis space M o l e c u l e s space o f space n i t r o g e n space g a s space i n space c y l i n d e r equals 24 cross times 10 to the power of 24 A v o g a d r o apostrophe s space n u m b e r space equals 6 cross times 10 to the power of 23 left parenthesis 1 right parenthesis M a s s space o f space n i t r o g e n space i n space a c y l i n d e r space equals fraction numerator 24 cross times 10 to the power of 24 cross times 28 over denominator 6 cross times 10 to the power of 23 equals 1120 g end fraction left parenthesis 2 right parenthesis V o l u m e space o f space n i t r o g e b space a t space S T P v o l u m e space o f space 28 g space o f space N subscript 2 equals 22.4 d m cubed v o l u m e space o f space 1120 g space o f space N subscript 2 equals fraction numerator 1120 cross times 22.4 over denominator 28 end fraction equals 896 space d m cubed$
(ii) The given equation is

$N a C l + A g N O_{3} \to A g C l + N a N O_{3}$

since, 143g of AgCl is formed from 58g of NaCl

1g AgCl will be formed from 58/143g of NaCl

14.3g of AgCl will be formed from 58/143 x 14.3g=5.8g of NaCl

% of NaCl = 5.8/30 x100= 19.33%

(iii)
$0.5 g space o c c u p i e s space 100 space c m cubed space 22400 c m to the power of 4 w i l l space w e i g h equals fraction numerator 0.5 over denominator 100 end fraction cross times 22400 equals 112 g s o space r e l a t i v e space m o l e c u l a r space m a s s space i s space 112 g$

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Similar questions

(a) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm³ of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below:
2C₂H₂_(g)+ 5O₂_(g) → 4CO₂_(g) + 2H₂O_(g)

(b) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37.
[N = 14, H = 1].

Calcium carbide is used for the artificial ripening of fruits. Actually, the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During the reaction, calcium hydroxide and acetylene gas is formed.

If 200 $c m^{3}$ of acetylene is formed from a certain mass of calcium carbide. Find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below:

$2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)$

A gas cylinder of capacity of 20 dm³ is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:

A gas cylinder contains 24 x 10²⁴ molecules of nitrogen gas. If Avogadro’s number is 6 x 10²³ and the relative atomic mass of nitrogen is 14, calculate:

200 c c

of acetylene is formed form a certain mass of calcium carbide. Find the volume of oxygen required and carbon dioxide formed during its complete combustion. The combustion reaction can be represented as:

2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)

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Element	% weight	Atomic weight	Atomic ratio	simple ratio
N	87.5	14	87.5/14=6.25	1
H	12.5	1	12.5/1=12.5	2