作业帮例2 将6.2gNa2O溶于93.8g水配成溶液(此时溶液的密度为1.2g/cm3),求CB?例5 已知KNO3在30度溶解度为70g,此时KNO3的饱和溶液的密度为1.6g/cm3,求溶液的CB?
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例2 将6.2gNa2O溶于93.8g水配成溶液(此时溶液的密度为1.2g/cm3),求CB?例5 已知KNO3在30度溶解度为70g,此时KNO3的饱和溶液的密度为1.6g/cm3,求溶液的CB?
例2 将6.2gNa2O溶于93.8g水配成溶液(此时溶液的密度为1.2g/cm3),求CB?
例5 已知KNO3在30度溶解度为70g,此时KNO3的饱和溶液的密度为1.6g/cm3,求溶液的CB?
例2 将6.2gNa2O溶于93.8g水配成溶液(此时溶液的密度为1.2g/cm3),求CB?例5 已知KNO3在30度溶解度为70g,此时KNO3的饱和溶液的密度为1.6g/cm3,求溶液的CB?
(1)Na2O + H2O = 2NaOH
62g 2mol
6.2g (0.2mol)
V = ( 6.2g + 93.8g ) / ( 1.2g/mL ) = 83.3mL =0.0833L
c(NaOH) = m(NaOH) / V = 0.2mol / 0.0833L = 2.4 mol/L
例5 ω（KNO3)= [70 g /( 100g + 70g )]*100% = 41.2%
c(KNO3) = 1000ρω/M =1000mL/L * 1.6 g/mL * 41.2% / 101 g/mol = 6.52 mol/L
反应得到氢氧化钠的物质的量是n=2×6.2/62=0.2mol 质量是m=8g 火碱的质量分数是ω=8%，那么c（NaOH）=1000ρω/M=1000×1.2×8%/40=2.4mol/L
c（KNO3）=1000ρω/M=1000×1.6×30/（30+70）/101=4.752mol/L
n=6.2/62=0.1mol.m=93.8+6.2=100g.体积V=100/1.2ml
所以CB=n/v=0.1molX1.2/100ml=1.2mol/L
(1)Na2O + H2O = 2NaOH
62g 80g
6.2g 8g
n(NaOH)=8g/40g/mol=0.2mol
V = (6.2g+93.8g) / (1.2g/cm3) = 83.3cm3 =0.0833L
c(NaOH) =...
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(1)Na2O + H2O = 2NaOH
62g 80g
6.2g 8g
n(NaOH)=8g/40g/mol=0.2mol
V = (6.2g+93.8g) / (1.2g/cm3) = 83.3cm3 =0.0833L
c(NaOH) = n(NaOH) / V = 0.2mol / 0.0833L = 2.4 mol/L
例5 ω（KNO3)= [70g /( 100g + 70g )]*100% = 41.2%
c(KNO3) = 1000ρω/M =1000mL/L * 1.6 g/mL * 41.2% /101 g/mol = 6.52 mol/L
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