Solution and Solvent Problems
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1. Calculate the molality of a solution of 14.0 g of Mn(NO3)2 in 250 g of water.
2. Calculate the molarity of a solution made by taking 60.0 mL of 2.79 M. NaOH and diluting it to a final volume of 260mL.
3. A natural alcohol has been isolated by an ethnobotanist working in the Amazon. A solution of 6.38 g of this alcohol, in 210 g of carbon tetrachloride, freezes at -29.6 degrees C. What is the molar mass of this substance? The freezing point depression constant of this solvent is 29.8 degrees C/m, while the freezing point is -22.3 degrees C.
4. A 15.9 %, by mass, solution of propanol, CH3CH2CH2OH, is found to have a density of 1.14 g/mL in water. What is the molarity of this solution, given these measurements?
5. Calculate the vapor pressure of water above a solution prepared, by adding 43.5 g. of lactose (C12H22O11) to 100.0 g. of water at 30.0 degrees C. The vapor pressure of water at this temperature is 31.8 torr.
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Solution Summary
In using the properties of molecules within their respective substances, the answers are determined through mathematical calculation and scientific knowledge of molecular structures and how they work. The relationships between the molecular structures of different substances help determine the number of molecules in each compound.
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1)
Molecular of Mn(NO3)2
= 55 + (14 + 16*3)*2
= 179 g
Therefore, the number of molecules of Mn(NO3)2 = 14.0/179
Hence, molality is the following:
m = number of molecules of Mn(NO3)2 / mass of water in kg
=> molality = (14.0/179)/0.25 = 0.312849 == 0.31 m --Answer
2)
60 ml of 2.79M NaOH
= 60*2.79/1000 molecules of NaOH
= 0.1674 molecules of NaOH
volume = 0.26 L
Therefore, molarity = 0.1674 molecules of NaOH/0.26 L = 0.644 M --Answer
3)
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- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India
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