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
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.

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, 

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
The violet light is the answer!
Answer:
Explanation — This page looks at the oxidation of alcohols using acidified sodium or ... of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. ... The electron-half-equation for this reaction is as follows: ... To do that, oxygen from an oxidizing agent is represented as [O]. ... Article type: Section or Page.
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