HARD

JEE Main

IMPORTANT

Earn 100

$500 mL$ of a hydrocarbon gas, burnt in excess of oxygen, yields $2500 mL$ of ${CO}_{2}$ and $3 L$ of water vapour, all volumes being measured at the same temperature and pressure. What is the formula of the hydrocarbon?

Important Questions on Eudiometry or Gas Analysis

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 7

When

0.02 L

of a mixture of hydrogen and oxygen was exploded,

0.003 L

of the oxygen remained. Calculate the initial composition of the mixture in percent by volume.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 8

12 mL

of a gaseous hydrocarbon was exploded with

50 mL

of oxygen. The volume measured after explosion was

32 mL

. After treatment with

KOH

, the volume diminished to

8 mL

. Determine the formula of the hydrocarbon.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 9

15 mL

of a gaseous hydrocarbon required

357 mL

of air

(21 %

of oxygen by volume) for complete combustion and the gaseous products occupied

327 mL

(all volumes being measured at NTP). What is the formula of the hydrocarbon?

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 10

0.90 g

of a solid organic compound (molecular weight

= 90 gm

) containing

C, H

and

O

was heated with oxygen corresponding to a volume of

224 mL

at STP. After the combustion, the total volume of the gases was

560 mL

at STP. On treatment with

KOH

, the volume decreased to

112 mL

. Determine the molecular formula of the compound.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 11

The explosion of a mixture containing one volume of a gas being studied and

1

volume of

H_{2}

yielded

1

volume of water vapor and

1

volume of

N_{2}

, with the volumes measured under identical conditions. Find the formula of the gas under study.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 12

5 mL

of a gas containing

C

and

H

was mixed with an excess of

O_{2}

(30 mL)

and the mixture exploded due to an electric spark. After the explosion, the volume of the mixed gases remaining was found to be

25 mL

. Adding a concentrated solution of

KOH

, the volume further diminished to

15 mL

, with the residual gas being pure oxygen. All volumes have been reduced to NTP. Determine the molecular formula of the hydrocarbon gas.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 13

40 mL

of ammonia gas taken in an eudiometer tube was subjected to sparks till the volume did not change any further. It was observed that the volume increased by

40 mL

40 mL

of oxygen was then mixed and the mixture was further exploded. The gases remained accounted to

30 mL

. Deduce the formula of ammonia.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 3 - Eudiometry or Gas Analysis>PROBLEMS>Q 14

20 mL

of a gas containing

H

and

S

was heated with tin. When the reaction was over, there was no change in volume. The residual gas was hydrogen. If the molecular weight of the gas is

34,

calculate the molecular formula.

500 mL of a hydrocarbon gas, burnt in excess of oxygen, yields 2500 mL of CO2 and 3 L of water vapour, all volumes being measured at the same temperature and pressure. What is the formula of the hydrocarbon?

Important Questions on Eudiometry or Gas Analysis

Learn and Prepare for any exam you want !

$500 mL$ of a hydrocarbon gas, burnt in excess of oxygen, yields $2500 mL$ of ${CO}_{2}$ and $3 L$ of water vapour, all volumes being measured at the same temperature and pressure. What is the formula of the hydrocarbon?