How many grams of nh3 are needed to provide the same number of molecules as in 0.75 g of sf6

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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