The depression in freezing point of 0.01 m aqueous CH 3 COOH solution is 0.02046° .1 m urea solution freezes at -1.86° C. Assuming molality equal to molarity. pH of CH 3 COOH solution is

Question

The depression in freezing point of 0.01 m aqueous CH 3 COOH solution is 0.02046° .1 m urea solution freezes at -1.86° C. Assuming molality equal to molarity. pH of CH 3 COOH solution is (A) 2 (B) 3 (C) 3.2 (D) 4.2

Tardigrade Admin · Accepted Answer

Δ T f =K f × m prime × i (Δ T( text urea )/Δ T( CH 3 COOH ))=(m prime text (urea) /m prime( CH 3 COOH ) × i) i=1.1(i=1 for urea) =(1+x) for CH 3 COOH x=0.1 ∴[ H ⊕]=C x=0.001 ∴ pH =3

Q. The depression in freezing point of $0.01 m$ aqueous $C H_{3} COO H$ solution is $0.0204 6^{\circ} .1 m$ urea solution freezes at $- 1.8 6^{\circ} C$ . Assuming molality equal to molarity. $p H$ of $C H_{3} COO H$ solution is

2

3

3.2

4.2

$Δ T_{f} = K_{f} \times m^{'} \times i$
$\frac{Δ T ( urea )}{Δ T ( C H _{3} COO H )} = \frac{m ^{'} (urea)}{m ^{'} ( C H _{3} COO H ) \times i}$
$i = 1.1 (i = 1$ for urea)
$= (1 + x)$ for $C H_{3} COO H$
$x = 0.1$
$∴ [H^{\oplus}] = C x = 0.001$
$∴ p H = 3$

Q. The depression in freezing point of 0.01m aqueous CH3​COOH solution is 0.02046∘.1m urea solution freezes at −1.86∘C. Assuming molality equal to molarity. pH of CH3​COOH solution is

2

3

3.2

4.2

ΔTf​=Kf​×m′×i ΔT(CH3​COOH)ΔT( urea )​=m′(CH3​COOH)×im′ (urea) ​ i=1.1(i=1 for urea) =(1+x) for CH3​COOH x=0.1 ∴[H⊕]=Cx=0.001 ∴pH=3

Q. The depression in freezing point of $0.01 m$ aqueous $C H_{3} COO H$ solution is $0.0204 6^{\circ} .1 m$ urea solution freezes at $- 1.8 6^{\circ} C$ . Assuming molality equal to molarity. $p H$ of $C H_{3} COO H$ solution is

$Δ T_{f} = K_{f} \times m^{'} \times i$
$\frac{Δ T ( urea )}{Δ T ( C H _{3} COO H )} = \frac{m ^{'} (urea)}{m ^{'} ( C H _{3} COO H ) \times i}$
$i = 1.1 (i = 1$ for urea)
$= (1 + x)$ for $C H_{3} COO H$
$x = 0.1$
$∴ [H^{\oplus}] = C x = 0.001$
$∴ p H = 3$