mass of K2SiO3 contains 2.50×10^23 atoms of K

mass of K2SiO3 contains 2.50×10^23 atoms of K

 

 

Determine the mass of potassium silicate that contains 2.50×10^23 atoms of potassium.

 

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> K2SiO3의 몰질량 = 154.28 g/mol

> K의 몰질량 = 39.10 g/mol

 

 

 

2.50×10^23 개 K × (1 mol / 6.022×10^23 개) × (39.10 g K / 1 mol K) × (154.28 g K2SiO3 / (2×39.10 g K))

= (2.50×10^23) × (1 / (6.022×10^23)) × (39.10) × (154.28 / (2×39.10))

= 32.0 g K2SiO3