how many moles of KF are contained in 347g of water in a 0.175 m KF solution?PLEASE HELP, have been struggling with the last past hour.?

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How many moles of KF are contained in 347 g of water in a 0.175 m KF solution? I would appreciate any assistance, as I have been struggling with this problem for the past hour. Thank you!

Joanny Hoeger · Accepted Answer

This is super late but [ 0.175 m KF] is the keyword here, the little 'm' means molality which is moles of solute/ kg of solvent. KF is the solute which we are trying to figure out, while the kg of solvent is the water (347g). 
Equation to work/plug in is  0.175 m KF= moles of KF/Kg of solvent (water)

Convert 347 g into kg, by multiplying 347 x 10^-3, which gives you .347 kg.
Plug kg of water into bottom of equation.

Solve for moles of KF by multiplying 0.175 m by .347 
(0.175m X .347 Kg) =.060725 moles of KF,

Answer: sig figs says= [6.07 x 10^-2]

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Anonymous · Answer

exactly 5.954

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how many moles of KF are contained in 347g of water in a 0.175 m KF solution?PLEASE HELP, have been struggling with the last past hour.?

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