Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]
Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: 
Molar mass of A = 
The depression in freezing point of solution with B solute: 
Molar mass of B = 
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
This means compound B has greater molar mass than compound A,
It is effected by diffusion (the power of smell and wind spread) but a solid is not.
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<h2>Answer:</h2>
<h2>Explanations</h2>
The complete balanced equation for the given reaction is expressed as;
Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane
According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:
Determine the mass of water produced
Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
