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
997 views
in Chemistry by (83.4k points)
closed by
A sample of pure compound contains 1.15 of sodium, `3.01xx10^(22)` atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is
A. `Na_(2)CO_(3)`
B. `NaCO_(2)`
C. `Na_(2)CO`
D. `Na_(2)CO_(2)`

1 Answer

0 votes
by (90.9k points)
selected by
 
Best answer
Correct Answer - B
Moles of sodium `= (1.15g)/(23gmol^(-1)) = 0.05 mol`
Moles of carbon `= (3.01xx10^(22))/(6.02xx10^(23)) = 0.05 mol`
Moles of oxygen = 0.1
Mole ratio, Na : C : O = 1:1:2
`:. E.F. = NaCO_(2)`

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

...