Answer:
Amount of NH3 in g = 0.0867 g
Explanation:
Molecular mass of [tex]SF_6 = 146.06\;g/mol[/tex]
[tex]Mole = \frac{0.75\;g}{146.06\;g/mol}=0.0051\;mol[/tex]
0.0051 mol has same no. of molecules in both NH3 and SF6.
Amount of NH3 in g = Mole × Molar mass
Molar mass of NH3 = 17.00 g/mol
Amount of NH3 in g = 0.0051 × 17.00
= 0.0867 g
0.087 g of NH₃ is needed to provide the same number of molecules as in 0.75 g of SF₆.
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ molecules. This implies that equal number of mole will contain the number of molecules.
To solve the question given above, we'll begin by calculating the number of mole in 0.75 g of SF₆. This can be obtained as follow:
Mass of SF₆ = 0.75 g
Molar mass of SF₆ = 32 + (19×6)
= 32 + 114
= 146 g/mol
Mole = mass / molar mass
Mole of SF₆ = 0.75 / 146
Thus, 5.14×10¯³ mole of SF₆ and 5.14×10¯³ mole NH₃ will contain the same number of molecules.
Finally, we shall determine the mass of 5.14×10¯³ mole of NH₃. This can be obtained as follow:
Mole of NH₃ = 5.14×10¯³ mole
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass = mole × molar mass
Mass of NH₃ = 5.14×10¯³ × 17
Therefore, 0.087 g of NH₃ is needed to provide the same number of molecules as in 0.75 g of SF₆.
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