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InterviewSolution
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What is molality of 1M solution of NaNo3 if it's density is 1.25g/cm3 |
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Answer» 1.00 L of the 1.00 M solution contains 1.00 mole of NaNO3 and this is 84.994 g. The total mass of the solution is this: 1000 cm^3 times 1.25 g/cm^3 = 1250 g How much water is in the solution: 1250 g minus 85.0 g = 1165 g molality: 1.00 mol / 1.165 kg = 0.86 m I rounded it to two sig fig, which seemed appropriate. You may include more if you wish. I also rounded off the 84.994 to 85 for convenience in subtraction |
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