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
–1 vote
175 views
in Equilibrium by (24 points)
The molar mass of a sparingly soluble salt \( \left( MX _{2}\right) \) is \( 100 g mol ^{-1} .0 .4 g \) of the salt is required to prepare a saturated solution with water at \( 25^{\circ} C \). Choose the correct statement(s) from the following the molar solubility product constant \( \left(K_{s p}\right) \) of the salt is \( 256 \times 10^{-9} mol L^{-1} \) the molarity of the saturated solution of the salt is \( 0.4 M \) the salt will precipitate as \( MX _{2} \) (s), if \( 500 mL \) of \( 10^{-4} M ^{\prime} MCl _{2} \) ' solution is mixed with \( 500 mL \) of \( 10^{-4} M \) ' \( NX \) ' solution. \( \left[ MCl _{2}\right. \) and \( NX \) are completely ionized in water \( ] \) The molarity of \( X \) ions in the saturated solution is \( 0.008 M \)

Please log in or register to answer this question.

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

...