Depression in freezing point of x m aqueous solution of propanoic acid is 0-02046- C- 0-1 molal glucose solution freezes at -0-186- C- Assuming molality of solution equal to molarity- If pH of propanoic acid is 3- then what will be the value of x-

For glucose solution, Δ T_f=K_f. m ⇒ K_f=0.1860.1=1.86 0.02046=(1+α )× 1.86 × x (1+α )=0.020461.86 x ...(1) Also, 10^13=x×α ∴ nbsp;α =10^ 3x nbsp;...(2) On ...

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Standard XII

Physics

Area and Volume Expansion

Depression in...

Question

Depression in freezing point of $x m$ aqueous solution of propanoic acid is ${0.02046}^{\circ} C . 0.1$ molal glucose solution freezes at $- {0.186}^{\circ} C .$ Assuming molality of solution equal to molarity. If $p H$ of propanoic acid is $3$ , then what will be the value of $x$ ?

0.22

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0.01

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0.15

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0.001

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Solution

The correct option is D $0.001$
For glucose solution,
$Δ T_{f} = K_{f} . m$
$\Rightarrow K_{f} = \frac{0.186}{0.1} = 1.86$
$0.02046 = (1 + α) \times 1.86 \times x$
$(1 + α) = \frac{0.02046}{1.86 x} . . . (1)$
Also, $10^{13} = x \times α$
$∴ α = \frac{10^{- 3}}{x} . . . (2)$
On solving equation (1) & (2), we get
$x = 0.01 M$

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