The quantity doesn't matter when it comes to boiling point, it's an Intensive property. The boiling point will not change. = 56°C
Density of the mixture = mass of the mixture / volume of the mixture
Mass of the mixture = mass of antifreeze solution + mass of water.
Mass of antifreeze solution = density of the antifreeze solution * volume
Mass of antifreeze solution = 0.8g/ml * 5.1 l * 1000 ml / l = 4,080 g
Mass of water = density of water * volume of water = 1.0 g/ml * 3.8 l * 1000 ml / l = 3,800 g
Mass of mixture = 4080 g + 3800 g= 7880 g
Volume of mixture = volume of antifreeze solution + volume of water
Volume of mixture = 5100 ml + 3800 ml = 8900 ml
Density of mixture = 7800 g / 8900 ml = 0.876 g/ml
Specific gravity of the mixture = density of the mixture / density of water = 0.876
Answer: 0.876
To Find :
Number of moles of C₃H₆O present in a sample weighing 25.6 grams.
Solution :
Molecular mass of C₃H₆O is :
M = (6×12) + (6×1) + (16×1) grams
M = 94 grams/mol
We know, number of moles of 25.6 grams of C₃H₆O is :
Hence, this is the required solution.
% H = 100 - ( 52.14 + 34.73 )=13.13 %
<span>assume 100 g of this compound </span>
<span>mass H = 13.13 g </span>
<span>moles H = 13.13 g / 1.008 g/mol=13 </span>
<span>mass C = 52.14 g </span>
<span>moles C = 52.14 g/ / 12.011 g/mol=4 </span>
<span>mass O = 34.73 g </span>
<span>moles O = 34.73 g/ 15.999 g/mol=2 </span>
<span>the empirical formula is C4H13O2</span>
