Chemistry How many moles of NH3 in the supplied solution0.2 M NH3 in 2 M Nh4NO3

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The discussion centers on calculating the moles of NH3 in a 2 ml solution containing 0.2 M NH3 and 2 M NH4NO3, with the answer required in mmol. The ratio of NH3 to NH4NO3 is established as 1:10, leading to a calculation of 0.182 ml for NH3. The final calculation yields 0.0364 mmol of NH3. There is confusion regarding the relevance of the ratio and the purpose of the 2/11 calculation. Clarity on these points is sought to ensure accurate understanding and application of the concepts.
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Homework Statement



How many moles of NH3 are in 2ml of supplied solution: 0.2 M NH3 in 2 M Nh4NO3

Ans to be given in mmol

Homework Equations



Concentration = mol/ L

The Attempt at a Solution



Ratio of NH3 : NH4NO3 = 1: 10

2 / 11 = 0.182 ml

0.2 mol/L * 1L/ 1000ml * 0.182 ml * 1000 = 0.0364 mmol
 
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everscern said:

Homework Statement



How many moles of NH3 are in 2ml of supplied solution: 0.2 M NH3 in 2 M Nh4NO3

Ans to be given in mmol

Homework Equations



Concentration = mol/ L

The Attempt at a Solution



Ratio of NH3 : NH4NO3 = 1: 10

2 / 11 = 0.182 ml

0.2 mol/L * 1L/ 1000ml * 0.182 ml * 1000 = 0.0364 mmol

I'm not sure why you're calculating the ratio of ammonia to ammonium salt, and then not using it? What's the 2/11 calculation for?
 
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