The freezing point depression of 0.001 m Kx[Fe(CN)6] is 7.10 × 10-3 K Determine the value of x. Given, Kf= 1.86 K kg mol-1 for water

Question

The freezing point depression of 0.001 m Kx[Fe(CN)6] is 7.10 × 10-3 K Determine the value of x. Given, Kf= 1.86 K kg mol-1 for water (A) 3 (B) 4 (C) 2 (D) 5

Tardigrade Admin · Accepted Answer

Δ x=i × Kf × m 7.10×10-3=i×1.86×0.001 i=3.817 α=(i-1/n-1) 1=(3.817-1/(x+1)-1) x=2.817=3 ∴ Molecular formula of the compound is K3 [Fe(CN)6]

Q. The freezing point depression of $0.001 m K_{x} [F e (CN)_{6}]$ is $7.10 \times 1 0^{- 3} K$ Determine the value of x. Given, $K_{f} = 1.86 K k g m o l^{- 1}$ for water

$3$

$4$

$2$

$5$

$Δ x = i \times K_{f} \times m$
$7.10 \times 1 0^{- 3} = i \times 1.86 \times 0.001$
$i = 3.817$
$α = \frac{i - 1}{n - 1}$
$1 = \frac{3.817 - 1}{( x + 1 ) - 1}$
$x = 2.817 = 3$
$∴$ Molecular formula of the compound is $K_{3} [F e (CN)_{6}]$

Q. The freezing point depression of 0.001mKx​[Fe(CN)6​] is 7.10×10−3K Determine the value of x. Given, Kf​=1.86Kkgmol−1 for water

3

4

2

5

Δx=i×Kf​×m 7.10×10−3=i×1.86×0.001 i=3.817 α=n−1i−1​ 1=(x+1)−13.817−1​ x=2.817=3 ∴ Molecular formula of the compound is K3​[Fe(CN)6​]

Q. The freezing point depression of $0.001 m K_{x} [F e (CN)_{6}]$ is $7.10 \times 1 0^{- 3} K$ Determine the value of x. Given, $K_{f} = 1.86 K k g m o l^{- 1}$ for water

$3$

$4$

$2$

$5$

$Δ x = i \times K_{f} \times m$
$7.10 \times 1 0^{- 3} = i \times 1.86 \times 0.001$
$i = 3.817$
$α = \frac{i - 1}{n - 1}$
$1 = \frac{3.817 - 1}{( x + 1 ) - 1}$
$x = 2.817 = 3$
$∴$ Molecular formula of the compound is $K_{3} [F e (CN)_{6}]$