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viva [34]
3 years ago
11

3. Could a molecule contain only one atom?

Chemistry
1 answer:
Blababa [14]3 years ago
8 0
A molecule is defined as two or more atoms of the same element different element that are bound together. A molecule may be homonuclear, which means, it consists of atoms of one chemical element, as with oxygen (O2); or it may be heteronuclear, a chemical compound composed of more than one element, as with water (H2O).
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What is the mole fraction of solute in a 3.19 m aqueous solution?
Bumek [7]
Molality= mol/ Kg

if we assume that we have 1 kg of water, we have 3.19 moles of solute. 

the formula for mole fraction --> mole fraction= mol of solule/ mol of solution

1) if we have 1 kg of water which is same as 1000 grams of water. 

2) we need to convert grams to moles using the molar mass of water 

molar mass of H₂O= (2 x 1.01) + 16.0 = 18.02 g/mol

1000 g (1 mol/ 18.02 grams)= 55.5 mol

3) mole of solution= 55.5 moles + 3.19 moles= 58.7 moles of solution

4) mole fraction= 3.19 / 58.7= 0.0543 
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3 years ago
8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a
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212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

C_{1} V_{1} =C_{2} V_{2}

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, (5.45*V_{1} ) = (820.7*1.41)

V_{1}=\frac{820.7*1.41}{5.45}=  212.33 ml

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

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