(a)

(i) Calculate the volume of 320 of $S O_{2}$ at STP. (Atomic mass : S = 32 and O = 16).
(ii) State Gay-Lussac's law of combining volume
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane $(C_{3} H_{8})$ (Atomic mass; C = 12, 0 = 16, H = 1, Molar volume $- 22.4 d m^{3}$

(b)

(i) An organic compound with vapour density = 94 contains
C = 12.67% H = 2.13 % and Br = 85.11% Find the molecular formula.
[Atomic mass ; C = 12, H = 1, Br = 80]
(ii) Calculate the mass of
(1) $10^{22}$ Atoms of sulphur.
(2) 0.1 mole of carbon dioxide.
[Atomic mass; S = 32, C = 12 and O = 16 and Avogadro's number $= 6 \times 10^{23}$

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Solution

$left parenthesis a right parenthesis left parenthesis i right parenthesis A t o m i c space m a s s space S equals 32 space a n d space O equals 16 M o l e c u l a r space m a s s space o f space S O subscript 2 equals 32 plus left parenthesis 2 cross times 16 right parenthesis equals 64 g A s space 64 space g space o f space S O subscript 2 space equals 22.4 space d m cubed T h e n comma space 320 space g space o f space S O subscript 2 equals fraction numerator 320 cross times 22.4 over denominator 64 end fraction equals 112 l i t r e s left parenthesis ii right parenthesis The space law space states space that long dash Under space same space conditions space of space temperature space and space pressure comma space the space volume space of space gases space taking space part space in a space chemical space reaction space show space s imple space whole space number space ratio space to space one space anothe space r and space to space those space of space products space if space gaseous. left parenthesis iii right parenthesis C subscript 3 H subscript 8 plus 5 O subscript 2 rightwards arrow 3 C O subscript 2 plus 4 H subscript 2 O M o l a r space m a s s space o f space p r o p a n e space equals 44 44 g space o f space p r o p a n e space r e q u i r e s space 5 cross times 22.4 space l i t r e space o f space o x y g e n space a t space S T P 8.8 g space o f space p r o p a n e space r e q u i r e s equals fraction numerator 5 cross times 22.4 cross times 8.8 over denominator 44 end fraction equals 22.4 space l i t r e s$

(b)
(i)

Element Relative atomic mass %compound atomic ratio Simple ratio

H 1 2.13 $\frac{2.13}{1} = 2.13$ 2

C 12 12.67 $\frac{12.67}{12} = 1.055$ 1

Br 80 85.11 $\frac{85.11}{80} = 1.063$ 1

Empirical formula =
n $\times$ (Empirical formula mass of ) = Molecular mass [2 $\times$ V.D]

n(12+2+80) = 94 $\times$ 2
n = 2
molecular formula =2 $\times$ Empirical formula
=

$left parenthesis ii right parenthesis left parenthesis 1 right parenthesis 10 to the power of 22 space end exponent a t o m s space o f space s u l p h u r space 6.022 cross times 10 to the power of 23 space a t o m s space o f space s u l p h u r space w i l l space h a v e space m a s s space equals 32 g 10 to the power of 22 space end exponent space a t o m s space o f space s u l p h u r space w i l l space h a v e space m a s s space equals fraction numerator 32 cross times 10 to the power of 22 space end exponent over denominator 6.022 cross times 10 to the power of 23 space end exponent end fraction equals 0.533 g left parenthesis 2 right parenthesis 0.1 space m o l e space o f space C O subscript 2 1 space m o l e space o f space C O subscript 2 space w i l l space h a v e space m a s s space equals 44 g 0.1 space m o l e space o f space C O subscript 2 space w i l l space h a v e space m a s s space equals 4.4 g$

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Similar questions

(i) Calculate the volume of 320 g of SO₂ at stp (Atomic mass: S = 32 and O = 16)
(ii) State Gay-Lussac's Law of combining volumes.
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈). (Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm³ at stp)

Calculate the mass of
(i) 10²² atoms of sulphur,
(ii) 0.1 mole of carbon dioxide.

[Atomic mass: S = 32, C = 12 and O = 16 and Avogadro's Number = 6 × 10²³]

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula.
[Atomic mass: C = 12, H = 1, Br = 80]

Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈).

[Atomic mass: C = 12, O = 16, H = 1, Molar volume = 22.4 dm³ at stp]

5.20 litres

3.50 litres

2.58 litres

4.48 litres

1.99 litres

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13% and Br = 85.11%. Find the molecular formula.

[Atomic mass: C = 12, H = 1, Br = 80]

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Element	Relative atomic mass	%compound	atomic ratio	Simple ratio
H	1	2.13	$\frac{2.13}{1} = 2.13$	2
C	12	12.67	$\frac{12.67}{12} = 1.055$	1
Br	80	85.11	$\frac{85.11}{80} = 1.063$	1