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
21.0k views
in Chemistry by (71.1k points)

The solubility product of chalk is 9.3 x 10–8. Calculate its solubility in gram per litre.

(a) 0.3 gram/litre 

(b) 0.0304 gram/litre

(c) 0.00304 gram/litre 

(d) None of these

1 Answer

0 votes
by (63.7k points)
selected by
 
Best answer

Correct option: (b) 0.0304 gram/litre

Explanation: 

CaCO3 ⇌ Ca2+ + CO32-

Let the solubility of CaCO3 be S mole per litre

Solubility in g/l = Mole/litre x Mol. wt of CaCO3

= 0.000304 x 100

= 0.0304 gram/litre

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

...