the maximum number of molecules is present in.
1) 5L of N2 gas at stp.
2) 0.5g of H2 gas.
3) 10 g of O2 gas.
4) 15 L if H2 gas at stp
Answers
Explanation:
no. of moles= Wt./M.Wt.-----1
at STP
no. of moles = V/22.4------2
no. of molecules = molesx NA(6.022x10^23)-------3
1) Present at STP
so use 2nd formula....
moles=5/22.4=0.22
by 3rd formula....
no. of molecules=0.22NA
2) it is not on STP so we will generally use 1st formula....
moles=0.5/2=0.25
no. of molecules=0.25NA
3)Its also not on STP
so applying 1st formula again.
moles=10/32=0.31
no. of moleculea=0.31NA
4) Now, its present on STP so use 2nd formula again dear....
moles=15/22.4=0.66
no.of molecules=0.66NA
Now you can see the ans. is clear...as no. of molecules are maximum in 15L of H2 gas at STP
Thankuu
hey mate their is your answer...
Dear Student,
Dear Student,Please find below the solution to the asked query:
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles(ii) Number of moles in 5 L of N2 gas at STP. = (5 L / 22.4 L) = 0.22 moles
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles(ii) Number of moles in 5 L of N2 gas at STP. = (5 L / 22.4 L) = 0.22 moles(iii) Number of moles in 0.5 g of H2 gas = (0.5 g / 2 g mol-1) = 0.25 moles
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles(ii) Number of moles in 5 L of N2 gas at STP. = (5 L / 22.4 L) = 0.22 moles(iii) Number of moles in 0.5 g of H2 gas = (0.5 g / 2 g mol-1) = 0.25 moles(iv) Number of moles in 10 g of O2 gas = (10 g / 32 g mol-1) = 0.312 moles
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles(ii) Number of moles in 5 L of N2 gas at STP. = (5 L / 22.4 L) = 0.22 moles(iii) Number of moles in 0.5 g of H2 gas = (0.5 g / 2 g mol-1) = 0.25 moles(iv) Number of moles in 10 g of O2 gas = (10 g / 32 g mol-1) = 0.312 molesHence, as we know that 1 mole of any gas has Avogadro number of molecules (6.022 x 1023) in it, we can say that the higher number of moles i.e. in 15 L of H2 gas will contain maximum number of molecules, as it has higher number of molar quantity.
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles(ii) Number of moles in 5 L of N2 gas at STP. = (5 L / 22.4 L) = 0.22 moles(iii) Number of moles in 0.5 g of H2 gas = (0.5 g / 2 g mol-1) = 0.25 moles(iv) Number of moles in 10 g of O2 gas = (10 g / 32 g mol-1) = 0.312 molesHence, as we know that 1 mole of any gas has Avogadro number of molecules (6.022 x 1023) in it, we can say that the higher number of moles i.e. in 15 L of H2 gas will contain maximum number of molecules, as it has higher number of molar quantity.Hope this information will clear your doubts about Some Basic Concepts of Chemistry.
Dear Student,Please find below the solution to the asked query:We can normalise all to one unit i.e. in terms of moles, from which we can say that whichever has greater molar quantity will contain greater number of molecules in it.(i) Number of moles in 15 L of H2 gas at STP. = (15 L / 22.4 L) = 0.67 moles(ii) Number of moles in 5 L of N2 gas at STP. = (5 L / 22.4 L) = 0.22 moles(iii) Number of moles in 0.5 g of H2 gas = (0.5 g / 2 g mol-1) = 0.25 moles(iv) Number of moles in 10 g of O2 gas = (10 g / 32 g mol-1) = 0.312 molesHence, as we know that 1 mole of any gas has Avogadro number of molecules (6.022 x 1023) in it, we can say that the higher number of moles i.e. in 15 L of H2 gas will contain maximum number of molecules, as it has higher number of molar quantity.Hope this information will clear your doubts about Some Basic Concepts of Chemistry.If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.