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blondinia [14]
3 years ago
5

Calculate the volumetric size of a water molecule in water vapor at normal conditions, assuming 1 mole of the vapor occupies 22.

4 l, as if the vapor were an ideal gas. Give answer in angstroms, two significant digits. Do not write down units in your answer.Calculate the volumetric size of a water molecule in water vapor at normal conditions, assuming 1 mole of the vapor occupies 22.4 l, as if the vapor were an ideal gas. Give answer in angstroms, two significant digits. Do not write down units in your answer.
Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
8 0

Let volumetric size of a water molecule = v

Since, 1 mole of water consists of 6.023\times 10^{23} molecules of water.

Thus, total volume of 1 mole of water = volumetric size \times 6.023\times 10^{23}

Substitute the value of total volume of 1 mole of water i.e. 22.4 L in above formula.

22.4 L = v\times 6.023\times 10^{23}

v= \frac{22.4 L}{6.023\times 10^{23}}

Convert the unit litre to angstrom

v=\frac{22.4 L}{6.023\times 10^{23}}\times \frac{1dm^{3}}{1L}\times \frac{(10^{9})^{3}(A^{o})^{3}}{1 dm^{3}}

= 3.7\times 10^{-23}\times 10^{27} (A^{o})^{3}

= 3.7 \times 10^{4}(A^{o})^{3}

Therefore, the volumetric size of the water molecule is 3.7 \times 10^{4}(A^{o})^{3}


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Answer : The mass of calcium chloride is, 116.84 grams

Solution : Given,

Molar mass of calcium chloride, CaCl_2 = 110.98 g/mole

Number of molecules of calcium chloride = 6.34\times 10^{23}

As we know that,

1 mole of calcium chloride contains 6.022\times 10^{23} molecules of calcium chloride

or,

1 mole of calcium chloride contains 110.98 grams of calcium chloride

Or, we can say that

As, 6.022\times 10^{23} molecules of calcium chloride present in 110.98 grams of calcium chloride

So, 6.34\times 10^{23} molecules of calcium chloride present in \frac{6.34\times 10^{23}}{6.022\times 10^{23}}\times 110.98=116.84grams of calcium chloride

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