In an experiment on depression of freezing point, two beakers leveled as A and B

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Question - In an experiment on depression of freezing point, two beakers leveled as A and B were filled with 1000 g. of water each. 40 g and 80 g of a certain non-electrolytic solute were added to both the beakers and freezing point of the solutions were noted as

T_{f A}

and

T_{f B}

respectively. Now both the solutions were poured into an empty beaker C and the freezing point of the mixed solution is found to be

T_{f C}

. What is the ratio

T_{f A} : T_{f B} : T_{f C} ?

Options:

A . 1 : 2 : 3

B . 2 : 4 : 3

C . 2 : 4 : 8

D . 1 : 4 : 8

Answer: Option B
:
B

△ T_{f} = K_{f} m

\Rightarrow 0 - T_{f} = K_{f} m

\Rightarrow= - K_{f} m

If 40 g of solute contains 'x'~moles of it then 80 g of the solut ill contains '2x' molesofit

m_{A} = \frac{x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{x}{\frac{1000}{1000}} = x

m_{B} = \frac{2 x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{x}{\frac{1000}{1000}} = 2 x

m_{C} = \frac{3 x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{3 x}{\frac{1000}{1000}} = 1.5 x

Required ration

T_{f A} : T_{f B} : T_{f c} = 1.2 : 1.5 = 2 : 4 : 3

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