JEE Main 2019 — Solutions Question with Solution

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

$$\frac{m_{pure}}{m_{dil}}=\frac{0.5}{0.2}$$ = $$\frac{5}{2}$$ ....(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 = $$\frac{n}{w_1}\times 1000$$ ....(4)

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

From (4) and (5), we get

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

Using equation (3), we get

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

$$\Rightarrow$$ 5w₁ = 2w₁ + 2w₀

$$\Rightarrow$$ 3w₁ = 2w₀

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

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