Explanation:
Relation between temperature and molality is as follows.
[tex]\Delta T_{f} = k_{f}m[/tex]
where, [tex]\Delta T_{f}[/tex] = depression in freeziing point
[tex]k_{f}[/tex] = molal deression in freezing point
m = molality
Putting the given values into the above formula as follows.
[tex]\Delta T_{f} = k_{f}m[/tex]
= [tex]1.86^{o}C/m \times 2.9 m[/tex]
= [tex]5.394^{o}C[/tex]
It is known that freezing point of water is [tex]0^{o}C[/tex] and the frreezing point of sucrose is calculated as follows.
Freezing point = [tex]0^{o}C - 5.4^{o}C[/tex]
= [tex]-5.4^{o}C[/tex]
Thus, we can conclude that the freezing point of a 2.9 m aqueous sucrose solution is [tex]-5.4^{o}C[/tex].
Given:
As we know the formula,
Depression in freezing point,
→ [tex]\Delta T_f = K_f \ m[/tex]
By substituting the values, we get
[tex]= 1.86\times 2.9[/tex]
[tex]= 5.4^{\circ} C[/tex]
hence,
The freezing point of sucrose solution will be:
= [tex]0^{\circ} C- 5.4^{\circ}C[/tex]
= [tex]-5.4^{\circ} C[/tex]
Thus the above response is right.
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