Question #65343
How many grams of ammonium nitrate need to be added to 500g of water so that the freezing point becomes -8.3degrees celcius?
Expert's answer
Answer
T=km
T=8,3
k=1,84 (constant for water)
m – molality
8,3/1,84=m=m(NH4NO3)/(M(NH4NO3)*0,5)
m(NH4NO3)=8,3*80*0,5/1,84=180 g
T=km
T=8,3
k=1,84 (constant for water)
m – molality
8,3/1,84=m=m(NH4NO3)/(M(NH4NO3)*0,5)
m(NH4NO3)=8,3*80*0,5/1,84=180 g
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