JEE Main 2024 · ChemistryNumerical answerNumerical valueEasyCalculation
JEE Main 30 January 2024, Shift 1, Chemistry Q81
Question 81 of 90 in this shift, Chemistry question 21 of 30, Section B.
The mass of sodium acetate () required to prepare 250 mL of 0.35 M aqueous solution is ________ g. (Molar mass of is 82.02 g mol)
Official answer
7
NTA final key.
- Topic
- Mole Concept
- Idea tested
- Molarity Calculation