QJEEM19Ch537d: JEE Main 2019 Chemistry - Class 12 JEEM19Ch537d Solution: The freezing point of a diluted milk sample is found to be –0.2 …

We know,

$\Delta$ T_f = i $\times$ k_f $\times$ m

$\therefore$ $\Delta$ T_{fDil. Milk} = (1) $\times$ k_f $\times$ m_dil = 0.2 ...(1)

$\Delta$ T_{fPure Milk} = (1) $\times$ k_f $\times$ m_pure = 0.5 ...(2)

From (1) and (2), we get

${{{m_{pure}}} \over {{m_{dil}}}} = {{0.5} \over {0.2}}$ = ${5 \over 2}$ ....(3)

Assume moles of fat in both the milk samples = n

Weight of milk = w₁

and weight of water = w₀

Molality of pure milk,
m_pure = ${n \over {{w_1}}} \times 1000$ ....(4)

Molality of diluted milk,
m_dil. = ${n \over {{w_1} + {w_0}}} \times 1000$ ....(5)

From (4) and (5), we get

${{{m_{pure}}} \over {{m_{dil}}}} = {{{w_1} + {w_0}} \over {{w_1}}}$

Using equation (3), we get

${5 \over 2} = {{{w_1} + {w_0}} \over {{w_1}}}$

$\Rightarrow$ 5w₁ = 2w₁ + 2w₀

$\Rightarrow$ 3w₁ = 2w₀

$\Rightarrow$ ${{{w_1}} \over {{w_0}}} = {2 \over 3}$

Which means 3 cups of water has been added to 2 cups of pure milk.