What Is the Molality of a 1.00 M LiBF4 Solution with a Density of 0.852 g/mL?

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To find the molality of a 1.00 M LiBF4 solution with a density of 0.852 g/mL, first calculate the mass of 1 liter of the solution, which is 852 grams. Next, determine the number of moles of LiBF4 in the solution, which is 1 mole since it is a 1.00 M solution. The mass of LiBF4 (molar mass approximately 65.39 g/mol) is about 65.39 grams, leaving approximately 786.61 grams of solvent (acetonitrile). Finally, convert the mass of the solvent to kilograms and use the formula for molality (moles of solute per kg of solvent) to find the molality. The final result will provide the required molality of the solution.
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Homework Statement


hey guys just wondering how to do this.
Lithium tetrafluoroborate (LiBF4) is sold as a 1.00 M solution in
acetonitrile. If the solution has a density of 0.852 g/mL,
the molality of the solution is?


i tired using denisty=m/v and concentration=n/v but always had 2 variables. any help would be good. thanks
 
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1 L of your solution - what mass? What mass of LiBF4? What mass of solvent?
 
umm, 852 grams/litre?
 
Any doubts?
 
i think it right but don't know what the next step is?
 
Perhaps if you will read my very first post it will become clearer.
 
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