Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a lower temperature than a solution containing 5.00 g of compound B in 100. g of water. Which compound has the greater molar mass

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,

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

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