$K_{s p}$ of $B a C r O_{4}$ is $2 \times 10^{- 10}$ at room temperature. Calculate the mass of Barium chromate lost by washing with three litres of water. Will the loss in mass depend on the quantity of precipitate?

5.64 mg

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10.625 mg

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3.88 mg

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15.6 mg

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Solution

The correct option is B
10.625 mg

$B a C r O_{4} (s)$ $⇌$ $B a^{2 +} + C r O_{4}^{2 -}$

$K_{S P}$ = $2 \times 10^{- 10}$

$∴$ Solubility of $B a C r O_{4}$ = $\sqrt{2 \times 10^{- 19}}$ = $1.41 \times 10^{- 3}$

Concentration of barium chromate = $1.4 \times 10^{- 3}$

M. Mass of $B a C r O_{4}$ = 253

$1.4 \times 10^{- 7}$ moles of $B a C r O_{4}$ = $1.4 \times 10^{- 3} \times 253 g$

= $1.4 \times 10^{- 3} \times 253 \times 1000$ = $3.542 m g$

for a single washing with one litre of water the mass of $B a C r O_{4}$ lost = $3.542 m g$

mass of Barium Chromate lost by washing with three litres = $3.542 \times 3$ = $10.626 m g$

Loss of mass does not depend upon the quantity of the precipitate.

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Q. Consider the following solubility data for various chromates at 25

^{o}

K_{s p}

A g_{2} C r O_{4}

9.0 \times 10^{- 12}

B a C r O_{4}

2.0 \times 10^{- 10}

P b C r O_{4}

1.8 \times 10^{- 14}

The chromate that is the most soluble in water at 25

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Consider the following solubility data for various chromates at 25 °C.

The chromate that is the most soluble in water at 25 °C on a molar basis is:

Q. Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of

1.32 g

. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of

0.0825 g

.
Calculate the relative molecular mass of oxygen.

When heated, potassium permanganate decomposes according to the following equation:
2KMnO₄ $\overset{}{\to}$ K₂MnO₄ + MnO₂ + O₂
(a) Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres). (K = 39, Mn = 55, O = 16)

$K M n O_{4}$ decomposes on heating according to the equation-

$2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$

On heating $K M n O_{4}$ , 1 litre of oxygen was collected at room temperature and it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g. Calculate the relative molecular mass of oxygen.

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Ksp of BaCrO4 is 2 × 10−10 at room temperature. Calculate the mass of Barium chromate lost by washing with three litres of water. Will the loss in mass depend on the quantity of precipitate?

$K_{s p}$ of $B a C r O_{4}$ is $2 \times 10^{- 10}$ at room temperature. Calculate the mass of Barium chromate lost by washing with three litres of water. Will the loss in mass depend on the quantity of precipitate?