Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
65 views
in Chemistry by (93.9k points)
closed by
An element crystallizes in b.c.c. lattice with cell edge of 500 pm. The density elements is 7.5g cm-3. How many atoms are present in 300 g of the element ?

1 Answer

0 votes
by (91.1k points)
selected by
 
Best answer
`f =(ZM)/(a^(3) xx Na)["Z for bcc"=2]`
`7.5 = (2 xx M)/((500 xx 10^(-10))^(3) xx 6.022 xx 10^(23))`
`M = (7.5 xx 500 xx 500 xx 500 xx 10^(-30) xx 6.022 xx 10^(23))/(2)`

M = 282.3 g/mol.
282.3 g of element contains `=6.022 xx 10^(23)` atoms.
300 g " " " " `=(6.022 xx 10^(23))/(282.3) xx 300 = 6.399 xx 10^(23)` atoms

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...