Home
Class 11
CHEMISTRY
The weight of a gaseous mixture containi...

The weight of a gaseous mixture containing `12.044xx10^(23)` atoms of He and `3.011xx10^(23)` molecules of hydrogen is _____________g.

Text Solution

Verified by Experts

The correct Answer is:
9
Helium weight = `(12.044xx10^(23))/(6.023xx10^(23))xx4=8g`
Hydrogen weight = `(3.011xx10^(23))/(6.023xx10^(23))xx2=1g`
Total = `8+1=9g`
Promotional Banner

Similar Questions

Explore conceptually related problems

6.023xx10^(22) atoms of Hydrogen can make

1 gram of hydrogen contains 6xx10^(23) atoms. Then 4 grams of He contains

When we concept that one mole of a substance contains the same numbers of elementary entities or one mole of any other substance, we don’t actually need to know what that number is. Sometimes however we will need to work with the actual number of elementary entities in a mole of substance. This number is called Avogadro.s number. N_(A)=6.022137xx10^(23) molecules The unit mol^(-1) which we read as per mole signifies that a collection of N_(A) molecular level entities is equivalent to one mole at the macroscopic level. For example a mole of carbon contains 6.02xx10^(23) atoms of C. A mole of oxygen gas contains 6.02xx10^(23) molecules of O_(2) . The number of atoms present in 8 g of ozone is

How many liters of helium gas at NTP has weight same as 3.0115xx10^(23) molecules of methane?

10 g of hydrogen contain ……………… molecules. If 2g of hydrogen contain 6.02xx10^(23) molecules.

A mole of any substance contains 6.023xx10^(23) particles. The particles may be atom, molecule ions, electron, proton or neutron. One mole of atom is equal to 1 gm -atom which is equal to atomic weight of atom. 1 gm molecule of any gas is 1 mole of gas whose volume is 22.4 litre at N.T.P. The volume of 3.011xx10^(23) atoms of hydrogen gas at N.T.P. is

A cylinder contains 6.023 xx 10^23 molecules of hydrogen and 5 xx 6.023 xx 10^22 molecules of oxygen. The partial pressure of oxygen is