You visited us 2 times! Enjoying our articles? Unlock Full Access!

Conductance of Electrolytic Solution

Chromium III ...

Question

Chromium (III) ion forms many compounds with ammonia. To find the formula of one of these compounds, we titrate $N H_{3}$ in the compound with standardized acid.

$C r (N H_{3})_{x} C l_{3} (a q) (A) + x H C l \to x {N H_{4}}^{+} (a q) + C r^{3 +} (a q)$ $+ (x + 3) C l^{-} (a q)$

Assume that $40$ mL of $1.5 M$ $H C l$ is used to titrate $2.635$ g of $A$ . The value of $x$ is (as nearest integer) :

Open in App

Solution

Molar mass of $C r (N H_{3})_{x} C l_{3} = 52 + 17 x + 35.5 \times 3 = 158.5 + 17 x$

$N H_{3}$ in the complex is neutralised by $H C l$ .

Moles of $H C l$ required $= \frac{40 \times 1.5}{1000}$ $= 0.06$

Moles of $A =$ $\frac{2.635}{(158.5 + 17 x)}$

As number of moles of reactants are equal,
$\frac{0.06}{x} = \frac{2.635}{158.5 + 17 x}$

$⟹ x = 5.89$

So. the nearest integer is $6$ .

Suggest Corrections

Similar questions

N a_{2} C O_{3} (a q) + 2 H C l (a q) \to 2 N a C l (a q) + C O_{2} (g) + H_{2} O (l)

38.55

H C l

2.150

N a_{2} C O_{3}

H C l

y \times 10^{- 2} (m o l . L^{- 1})

y

C o {({N H}_{3})}_{x} {C l}_{2} (a q) + H C l \to {N H}_{4}^{\oplus} (a q) + {C O}^{2 +} (a q) + {C l}^{-} (a q)

3.96 g m

40 m l

2 M

H C l

0.490

H_{2} A

K O H

H_{2} A (a q) + 2 K O H (a q) \to K_{2} A (a q) + 2 H_{2} O (l)

100

0.1 M

K O H

C o C l_{3} .4 N H_{3}

A g N O_{3} (a q)

A g C l

H_{2} S 0_{4}

400 m L

0.2 M H_{2} S O_{4}

600 m L

0.1 M N a O H

x \times 10^{- 2} K

H^{+} (a q) + O H^{-} (a q) \to H_{2} O : Δ_{γ} H = - 57.1 {kJ mol}^{- 1}

Specific heat of H_{2} O = 4.18 {J k}^{- 1} g^{- 1}

Density of H_{2} O = 1.0 {g cm}^{- 3}

Join BYJU'S Learning Program