A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving some urea () in of . This solution boils at . Calculate the mass of that was dissolved. Round your answer to significant digits.

This question is incomplete, the complete question is;

A certain liquid X has a normal boiling point of 150.4 °C and a molar boiling point elevation constant kb is 0.60 °Ckgmol⁻¹.

A solution is prepared by dissolving some urea (NH22CO) in 750 g of X. This solution boils at 150.9 °C . Calculate the mass of urea that was dissolved. Round your answer to 3 significant digits.

the mass of urea that was dissolved is 37.5 g

Explanation:

Given the data in the question;

normal boiling point of X; Tb⁰ = 150.4 °C

boiling point of solution Tb = 150.9 °C

Change in boiling point Δt = Tb - Tb⁰  = 150.9 °C - 150.4 °C = 0.5 °C

Kb = 0.6 °C.kg.mol⁻¹

V = 750 g

Now, we know that

Δt = Kb × molality

so

0.5 = 0.6 × molality

molality = 0.5 / 0.6

molality = 0.833

we know that molar mass of urea is 60 g/mol

so

molality = mass × 1000 / molar mass × volume( g )

we substitute

0.833 = ( mass × 1000 ) / ( 60 × 750 )

0.833 = ( mass × 1000 ) / 45000

0.833 × 45000 = mass × 1000

mass = ( 0.833 × 45000 ) / 1000

mass = 37485 / 1000

mass = 37.485 ≈ 37.5 g   { 3 significance figure }

Therefore, the mass of urea that was dissolved is 37.5 g

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 !