how to find the freezing point of a 1 m sugar (aqueous) solution? and find the molar mass?

Question

if you can help me figure out the problem i have no idea how to work it out please help

Answer 1

Sugar is C12H22O11.

To find the molar mass, look on the periodic table. The molar mass for C is 12.01 g/mol, H is 1.01 g/mol, O is 16.00 g/mol. Therefore, the molar mass of C12H22O11 is 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol.

To find the freezing point,

ΔTf = i * Kf * m = 1* 1.86 deg C/m * 1m = 1.86 deg C

0 - 1.86 = -1.86 deg C

Answer 2

The freezing point of a solution is always lower than that of the solvent

The formula for that phenomenon called cyoscopia is

lowering of t dT = K * molarity * i

i is a correcting term for dissociating molecules

as sugar does not dissociate i =1

K = -1.86 °C by mole of solute

dT lowering of temperature

here molarity =1 so dT = -1.86°

And as water freezes at 0C

the solution freezes at -1.86°C

Answer 3

RE: answer elemental diagnosis? question: Sorbitol is a sweet substance cutting-edge in culmination and berries and frequently used as a sugar replace. An aqueous answer containing a million.00 g sorbitol in one hundred.0 g water is discovered to have a freezing element of -0.102 ranges Celsius. Elemental diagnosis exhibits that sorbitol includes 39.56%...

Answer 4

there is a simple formula for this

the change in the freezing point is proportional to the molality

hence dTf(read as delta Tf i.e. change in the freezing point)
=K*m
where K is the proportionality constant and =1.86 for water
m is the molality(which is 1 in the given problem)
therefore the depression in freezing point =1.86
hence the freezing point = 0 (i.e.freezing point of water)-1.86
= -1.86
molality(m) is the no. of moles per kg of solvent (defn.)
since molality is directly given there is no need to find molar mass
but from memory i can tell you
weight of sucros(C12H22O11)=342gms

weight of glucose (C6H12O6)=180gms