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Standard XII

Chemistry

Henry's Law

In an experim...

Question

In an experiment during the analysis of a carbon compound, 145 l of $H_{2} O$ was collected at 760 mm Hg pressure at $27^{\circ}$ C temperature. The mass of $H_{2} O$ is nearly.......

10 g

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12 g

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24 g

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6 g

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Solution

The correct option is B
12 g

PV = nRT, n = $\frac{P V}{R T}$ = $\frac{1 \times 145}{0.082 \times 300}$ = 5.8 $\approx$ 6 mole.

1 mole of $H_{2}$ =2 g of $H_{2}$

So, 6 mole of $H_{2}$ = 2 $\times$ 6 = 12g

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In an experiment during the analysis of a carbon compound, 145 l of H2O was collected at 760 mm Hg pressure at 27∘ C temperature. The mass of H2O is nearly.......

The correct option is B 12 g PV = nRT, n = PVRT = 1×1450.082×300 = 5.8 ≈ 6 mole. 1 mole of H2=2 g of H2 So, 6 mole of H2 = 2×6 = 12g

In an experiment during the analysis of a carbon compound, 145 l of $H_{2} O$ was collected at 760 mm Hg pressure at $27^{\circ}$ C temperature. The mass of $H_{2} O$ is nearly.......

The correct option is B
12 g

PV = nRT, n = $\frac{P V}{R T}$ = $\frac{1 \times 145}{0.082 \times 300}$ = 5.8 $\approx$ 6 mole.

1 mole of $H_{2}$ =2 g of $H_{2}$

So, 6 mole of $H_{2}$ = 2 $\times$ 6 = 12g