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Fedral regulations set an upper limit of 50 parts per millison (ppm) of NH_(3) in the air in a work environment ( that is , 50 mL NH_(3) per 10^(6) mL of air).The density of NH_(3)(g) at room temperature is 0.771 g/L .Air from a manufacturing operation was drawn through a solution containing 100 mL of 0.0105 M HCl . The NH_(3) reacts with HCl as follows: NH_(3)(aq)+HCl(aq)toNH_(4)Cl(aq) After drawing air through the acid solution for 10 minutes at a rate 10 litres/min the acid was titrated .The remaining acid required13.1 mL of 0.0588 M NaoH to reach the equivalence point . How many ppm of NH_(3) were in the air ? |
| Answer» SOLUTION :`61.6 PPM` | |