The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0mg Na+ per mL is…………. g. (Rounded off to the nearest integer)
[Given ; Atomic weight in g mol-1 -Na : 23; N : 14; O : 16]
The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0mg Na+ per mL is…………. g. (Rounded off to the nearest integer)
[Given ; Atomic weight in g mol-1 -Na : 23; N : 14; O : 16]
-
1 Answer
-
Mass of Na+ in 50 ml = 70 × 50 = 3500 mg
23000 mg of Na+ is present in 85000 mg of NaNO3 (1 mole NaNO3 contains 1 mole Na+)
mg Na+ will be present in
= 12934.78 mg
= 12.93478 gm
Similar Questions for you
In the medical entrance test NEET, there can be 1 to 3 questions from this chapter. Some year, the Chemistry section of NEET has only one question from this chapter and in some other years, there can be 3 questions.
The following are the key concepts of this chapter: Compound, Elements, Rules, Law of conservation of mass, Addition and Subtraction, Atomic Mass, Law of multiple proportions, and Molecular Mass.
As the name suggests, the first chapter of the NCERT Class 11 Chemistry introduces various basic concepts of chemistry, such as the definition and importance of chemistry, atomic matter and molecular masses, the mole concept, laws of chemical combination, empirical, stoichiometry, and molecular formulas. It also includes the concepts of molarity and molality.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers