Question

solubility of ammonium chloride at 293 kelvin is 37 gram what mass of Ammonium Chloride would be needed to make a saturated solution in 10 gram of water at 293 Kelvin​

Answer 1

Answer:

3.7 gram of Ammonium Chloride would be needed to make a saturated solution in 10 gram of water at 293 K

Explanation:

Solubility of Ammonium Chloride (NH₄Cl) at 293 K is 37 gram/100 mL

⇒ 37 gram of NH₄Cl is soluble in 100 mL of water

or, 37 gram of NH₄Cl is soluble in 100 gram of water  

(∵ density of water = 1g/ml)

∵ In 100 gram of water amount of NH₄Cl soluble = 37 gram

∴ In 1 gram of water, amount of NH₄Cl soluble = 37/100 gram

                                                                              = 0.37 gram

∴ In 10 gram of water, amount of NH₄Cl soluble = 0.37 × 10 = 3.7 gram

Therefore, 3.7 gram of Ammonium Chloride would be needed to make a saturated solution in 10 gram of water