Here you go! There are 0.9307 moles in 123.0 g of the compound. I solved this by using a fence post method. I calculated the number of grams in one mol of (NH4)2 SO4 and got 132.16.
I did this by finding the atomic mass of each element on the periodic table (my work is in the color blue for this step)
After that, i divided the given mass by the mass of one mol of the compound.
The answer is 0.9307 moles!! I hope this helped you! :))
<span>As we know that
1 cu cm H2O = 1 mL H2O = 1g H2O
now
Heat of fusion of water = 79.8 cal/g
and
Heat of vaporization of water = 540 cal/g
Atomic weight of water : H=1 O=16 H2O=18
now by calculating and putting values
65.5gH2O x 79.8cal/gH2O x 1gH2O/540cal = 9.68g H2O (steam)
9.68gH2O x 1molH2O/18gH2O x 22.4LH2O/1molH2O = 12.0 L H2O
hope it helps</span>
Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
