how many mole , molecules, and atom of phosphorus are contained in 92.93g of phosphorus , if the formula of its molecules is P4 ?
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Molecular mass (p4)= 4 *×31 = 124u
1 mole of phosphorus = 124g = 6.022×10^23 molecules of phosphorus.
Now,
124g of phosphorus contain 6.022×10^23.
Therefore,
92.9g of phosphorus will contain = 6.022×10^23×92.9/124
= 4.51×10^23 molecules.
1 molecule of (p4) contain 4 atoms of phosphorus.
Therefore,
4.51×10^23 molecules of phosphorus will contain phosphorus atoms = 4×4.51×10^23 = 1.804×10^24 atoms.
✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️
Molecular mass (p4)= 4 *×31 = 124u
1 mole of phosphorus = 124g = 6.022×10^23 molecules of phosphorus.
Now,
124g of phosphorus contain 6.022×10^23.
Therefore,
92.9g of phosphorus will contain = 6.022×10^23×92.9/124
= 4.51×10^23 molecules.
1 molecule of (p4) contain 4 atoms of phosphorus.
Therefore,
4.51×10^23 molecules of phosphorus will contain phosphorus atoms = 4×4.51×10^23 = 1.804×10^24 atoms.
✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️
prabhat4425:
wrong ans
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