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

Solubility of ammonium chloride at 293 kelvin is 37 gram what mass of

1.	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» 3.7 gram of Ammonium Chloride would be needed to make a saturated solution in 10 gram of waterExplanation:GivenSolubility of Ammonium Chloride (NH₄Cl) at 293 K is 37 gram/100 mLi.e. 37 gram of NH₄Cl is SOLUBLE in 100 mL of wateror, 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 gramTherefore, 3.7 gram of Ammonium Chloride would be needed to make a saturated solution in 10 gram of water