
Calculate the mass of $1.05 \times {10^{26}}$ atoms of Neon? Give steps.
Answer
558.6k+ views
Hint: Whenever you want to find out mass, moles or number of atoms, you should start your calculation by keeping three formulas as your main steps for finding answers. First is molar mass, it is mass in one mole, second is the Avogadro number $6.022 \times {10^{23}}\,atoms$ and third is to convert atoms into mass and molar mass.
Complete solution:
We know that neon is a noble gas and has a position just right most of the periodic table. If we can find out any relation in which mass is included we can find out the mass for a given amount of atoms. Let’s start checking with molar mass so, molar mass of neon is $20\,g\,mo{l^{ - 1}}$ it means that $1\,mole$ of neon contains $20\,g$ of mass.
Step-1: $1\,mole\,of\,Neon = 20\,g$ this information comes from its molar mass, $1\,mole$ of neon contains $20\,g$ of mass. Here we have two terms one is mole and other is gram.
Step-2: Next step is to find the number of moles using the given amount of atoms, as per here we know that there are $6.022 \times {10^{23}}\,atoms$ in $1\,mole$.
$1\,mole = \,6.022 \times {10^{23}}\,atoms$
Or this statement can be written as $6.022 \times {10^{23}}\,atoms$ number of atoms form $1\,mole$ it means now we have to calculate moles from given amount of number, so then how much moles get formed by $1.05 \times {10^{26}}\,atoms$ amount.
$1.05 \times {10^{26}}\,atoms \times \dfrac{{1\,mole}}{{6.023\, \times {{10}^{23}}\,atoms}}$
Step-3: In the above calculation atom get cancelled from both numerator and denominator and we get a numeric value as $174.4\,mole$ as shown below.
$1.05 \times {10^3}\, \times \dfrac{{1\,}}{{6.023\,\,}}mole$ = $174.4\,mole$
Step-4: As we found the number of moles and we know that by the definition of molar mass we have $20\,g$ in $1\,mole$.
$1\,mole = \,20\,g\,of\,Neon$
$174.4\,mole = \,(20\, \times 174.4)\,g\,of\,Neon$
$174.4\,mole = \,3488\,g\,of\,Neon$
Note:Make sure that the approach of these types of questions will be easy so that you can get it in your mind. While converting these moles into gram and atoms into mole or vice-versa, do the calculation very carefully because the only mistake in this question can be your calculation or approach. The approach here is just three main formulas that we should keep in mind. First is molar mass, it is mass in one mole, second is the Avogadro number $6.022 \times {10^{23}}\,atoms$ and third is to convert atoms into mass and molar mass.
Complete solution:
We know that neon is a noble gas and has a position just right most of the periodic table. If we can find out any relation in which mass is included we can find out the mass for a given amount of atoms. Let’s start checking with molar mass so, molar mass of neon is $20\,g\,mo{l^{ - 1}}$ it means that $1\,mole$ of neon contains $20\,g$ of mass.
Step-1: $1\,mole\,of\,Neon = 20\,g$ this information comes from its molar mass, $1\,mole$ of neon contains $20\,g$ of mass. Here we have two terms one is mole and other is gram.
Step-2: Next step is to find the number of moles using the given amount of atoms, as per here we know that there are $6.022 \times {10^{23}}\,atoms$ in $1\,mole$.
$1\,mole = \,6.022 \times {10^{23}}\,atoms$
Or this statement can be written as $6.022 \times {10^{23}}\,atoms$ number of atoms form $1\,mole$ it means now we have to calculate moles from given amount of number, so then how much moles get formed by $1.05 \times {10^{26}}\,atoms$ amount.
$1.05 \times {10^{26}}\,atoms \times \dfrac{{1\,mole}}{{6.023\, \times {{10}^{23}}\,atoms}}$
Step-3: In the above calculation atom get cancelled from both numerator and denominator and we get a numeric value as $174.4\,mole$ as shown below.
$1.05 \times {10^3}\, \times \dfrac{{1\,}}{{6.023\,\,}}mole$ = $174.4\,mole$
Step-4: As we found the number of moles and we know that by the definition of molar mass we have $20\,g$ in $1\,mole$.
$1\,mole = \,20\,g\,of\,Neon$
$174.4\,mole = \,(20\, \times 174.4)\,g\,of\,Neon$
$174.4\,mole = \,3488\,g\,of\,Neon$
Note:Make sure that the approach of these types of questions will be easy so that you can get it in your mind. While converting these moles into gram and atoms into mole or vice-versa, do the calculation very carefully because the only mistake in this question can be your calculation or approach. The approach here is just three main formulas that we should keep in mind. First is molar mass, it is mass in one mole, second is the Avogadro number $6.022 \times {10^{23}}\,atoms$ and third is to convert atoms into mass and molar mass.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

