Asked by John
A mass of 10g of sodium nitrate is dissolved in 10cm3 of water at 80 degrees Celsius. As the solution is cooled, at what temperature will a precipate first appear?
All Answers 1
Answered by
GPT-5 mini
AI
Step 1 — express the composition as g solute per 100 g water:
10 cm3 water ≈ 10 g, so 10 g NaNO3 in 10 g water = 100 g NaNO3 per 100 g water.
Step 2 — find the temperature at which NaNO3 solubility = 100 g per 100 g water.
Using standard solubility data for NaNO3 (≈92 g/100 g H2O at 20 °C and ≈109 g/100 g H2O at 40 °C), the solubility reaches 100 g/100 g H2O at about 30 °C.
Answer: a precipitate will first appear on cooling at roughly 30 °C (± a few degrees depending on the exact solubility data and the approximation 10 cm3 water = 10 g).
10 cm3 water ≈ 10 g, so 10 g NaNO3 in 10 g water = 100 g NaNO3 per 100 g water.
Step 2 — find the temperature at which NaNO3 solubility = 100 g per 100 g water.
Using standard solubility data for NaNO3 (≈92 g/100 g H2O at 20 °C and ≈109 g/100 g H2O at 40 °C), the solubility reaches 100 g/100 g H2O at about 30 °C.
Answer: a precipitate will first appear on cooling at roughly 30 °C (± a few degrees depending on the exact solubility data and the approximation 10 cm3 water = 10 g).
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.