(NH4)3PO4의 몰분율. 4.012 m ammonium phosphate

(NH4)3PO4의 몰분율. 4.012 m ammonium phosphate

 

 

Calculate the mole fraction of the solute of a 4.012 m aqueous solution of ammonium phosphate with a density, ρ = 1.056 g/mL.

 

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4.012 m = 4.012 mol 용질 / 1 kg 용매

( 몰랄농도 설명 https://ywpop.tistory.com/2667 )

 

 

1 kg = 1000 g 용매(H2O)의 몰수를 계산하면,

1000 g / (18.02 g/mol) = 55.494 mol H2O

( 계산 설명 https://ywpop.tistory.com/7738 )

 

 

용질의 몰분율을 계산하면,

4.012 mol / (4.012 + 55.494 mol) = 0.06742

( 몰분율 설명 https://ywpop.tistory.com/2659 )

 

 

답: 0.06742