Please show work on how you solved Sorry its side way but i cant seem to flip it

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Please show work on how you solved

Sorry its side way but i cant seem to flip it

alos these questions

1. Calculate the freezing point depression if 225.0 grams of isoprpyl alcohol, C3H7OH, are added to 7.20x10^2 grams of water

(water Kf, c/m 1.86)

2. calculate the molar mas of a substance if the addition of 53.0 grams of the substance to 3.80 x 10^2 g of ethanol (kb = 1.22 c/m)

elevated the boiling point of ethanol by 2.13c

3. calculate he molality of a solution made by adding 98.0 grams of sodium chlride to 3.50 x 10^2 mL of water

4. the salt solution in Q 3 has a boiling point elevation that is twice the value of solution of sugar with the same molality, Brifly explain why

Thank you so much i dont have much time to complete it im working on it now and would appreciate the extra help!

Find the accepted molar mass of vanillin (4-hydroxy-3-methoxybenzaldehyde) using a standard reference such as the CRC Handbook of Chemistry and Physics. calculate a percent difference for your molar mass determination.

Explanation / Answer

1)mass of water=720 grams

molar mass of isopropyl alcohol=60 g/mol

so number of moles of it=225/60

=3.75

so molality=3.75*1000/720

=5.208 m

so freezing point depression=1.86*5.208

=9.6875 oC

2)let the molar mass be m

so number of moles=53/m

molality=(53/m)*(1000/380)

so,

2.13=1.22*(53/m)*(1000/380)

or m=79.88 g/mol

3)number of moles of NaCl=98/58.5

=1.675

so, the molality=1.675*(1000/350)

=4.785 m

4)it is beacause of the fact that the vant hoff factor of NaCl is 2 since it ionises to give separate sodium ion and chloride ion whereas for sugar it is 1 since it does not dissociates in water.

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