Question

please solve this...

100 points

Answer 1

Hi friend,

Your answer is:-

1.The sample D contains the maximum number of molecules.

2.If the temperature and the pressure of gas A are kept constant, then the volume of A will get doubled.

3.If this ratio of gas volumes refers to the reactants and products of a reaction,then this refers to Gay Lussac's law.

4.Ratio of volume of gases A and D= 1: 4   
Now, 4 × 5.6 dm3 at S. T. P. = 22.4 dm3 (Molar volume)
 = 6 × 10^23 molecules.

5.6 × 10^23 molecules is Avogadro’s number of molecules contained in 1 gram mole of the substance.

So, if gas D is said to be N2O then ,
1 gram mole of N2O = 2 × 14 + 16 = 44g.

Answer 2

(1)

we know ,
PV = nRT
if Pressure (P ) and temperature ( T) are constant then ,
V will be directly proportional to no of molecules ( n)

hence, no of molecules of gas is greater when volume of gas is greater .

so, " D " have maximum no of molecules .


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(2) T and P are constant then
volume will be directly proportional to no of moles or molecules . ( Avogadro's law )
hence if no of molecule is doubled then volume of gas also will be doubled .

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(3) Gay Lussac's Law of gaseous volumes :- becoz in a chemical reaction involing gases , are in simple ratio at constant temperature and pressure.

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(4) Volume of Gas A = 5.6 dm³ = 5.6L
no of moles = given volume /22.4 L
= 5.6 L/22.4 L
= 1/4

no of molecules = no of moles ×6 × 10²³
= 1/4 × 6 × 10²³ = 1.5 × 10²³

volume of Gas D = 4 × volume of Gas A
according to Avogadro's Law
no of molecules of Gas D = 4 × no of molecules of Gas A

no of molecules of Gas D = 6 × 10²³

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(v) no of moles = 6 × 10²³/6 × 10²³
= 1

mole = given wt/molecular wt
1 = given wt/ (14 × 2 + 16)
given wt = 44 g

hence , wt of Gas D = 44 g