Find Number Of Nitrogen Atoms Present In 65 Mg Of N A N 3 Atomic Mass Of N A Equal To 23 And Equal To 14 U
Answers
Answer:
Explanation:
Weight = W = 65 mg = 0.065g (Given)
Atomic Mass of NaN3 = 23 = 14U
Molar mass of NaN3 -
M = 23 + 3 × 14
M = 65 g/mol
Calculating the moles of NaN3 by the formula -
n = W / M
n = 0.065 / 65
n = 10^-3 mol
Thus, one mole of NaN3 contains 3 N-atoms, therefore 10^-3 moles NaN3 consists of -
N = 3 × 10^-3 × NA
N = 3 × 10^-3 × 6.022×10^23
N = 1.8066×10^21
Thus, 65 mg of NaN3 contains 1.8066×10^21 N-atoms.
Answer:
Explanation:
Weight = W = 65 mg = 0.065g (Given)
Atomic Mass of NaN3 = 23 = 14U
Molar mass of NaN3 -
M = 23 + 3 × 14
M = 65 g/mol
Calculating the moles of NaN3 by the formula -
n = W / M
n = 0.065 / 65
n = 10^-3 mol
Thus, one mole of NaN3 contains 3 N-atoms, therefore 10^-3 moles NaN3 consists of -
N = 3 × 10^-3 × NA
N = 3 × 10^-3 × 6.022×10^23
N = 1.8066×10^21
Thus, 65 mg of NaN3 contains 1.8066×10^21 N-atoms.