a sample of cabr2 and nabr contains 2.4×10^23 atoms of ca and 6.6×10^23 atoms of br. mass of nabr in the given sample is
Answers
Answered by
1
Explanation:
Moles of Na, n
Na
=
M
w
=
23
2.04
=0.088
Moles of C, n
C
=
N
A
N
=
6.022×10
23
2.65×10
22
=0.044
Moles of O, n
O
=0.132
Now,
n
Na
:n
C
:n
O
=0.088:0.044:0.132=2:1:3
The empirical formula of the compound =Na
2
CO
3
Answered by
5
Given:
A sample of CaBr₂ and NaBr contains 2.4 × 10²³ atoms of Ca and 6.6 × 10²³ atoms of Br.
To find:
Mass of NaBr.
Solution:
- Number of Br atoms in NaBr = 6.6 × 10²³
- Number of moles of Br atoms = Number of Br atoms ÷ Avogadro's number
- = 6.6 × 10²³/6 × 10²³
- = 1.1 moles.
- Number of Ca atoms in CaBr₂ = 2.4 × 10²³
- Number of moles of Ca atoms = Number of Ca atoms ÷ Avogadro's number
- = 2.4 × 10²³/6 × 10²³
- = 0.4 moles.
- CaBr₂ has two atoms of Br and one atom of Ca in one molecule.
- We can say that,
- Number of moles of Br in CaBr₂ = 2 × Number of moles of Ca in CaBr₂.
- = 2 × 0.4 = 0.8 moles.
- Number of moles of Br in NaBr = 1.1 - 0.8 = 0.3.
- Mass of NaBr = Number of moles of NaBr × Molar mass of NaBr
- = 0.3 × 103 = 30.9 grams.
Therefore, the mass of NaBr in the given sample is 30.9 grams.
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