Answer:
Mass of KNO3= 10g
Molar mass of KNO3 = 101.1032g/mol
Volume = 250ml = 0.25L
No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3
no of mole of KNO3 = 10/101.1032
No of mole of KNO3 = 0.09891
molarity of KNO3 = no of mole of KNO3/Vol (L)
Molarity = 0.09891/0.25 = 0.3956M
Molarity of KNO3 = 0.3956M
Empirical formula: The formula consist of proportions of the elements which is present in the compound or the simplest whole number ratios of atoms.
Now, molecular formula is equal to the product of n (ratio) and empirical formula.
Molecular formula = 
(1)
molecular formula =
(given)
Since, 6 is the smallest subscript in above molecular formula to get the simpler whole number of atoms. Therefore, divide all the subscripts i.e. number of carbon atoms (12), number of hydrogen atoms (24) and number of oxygen atoms (6) by 6.
empirical formula becomes 
Thus, according to the formula (1)
Hence, empirical formula of given molecular formula is
Answer is B. gas formation
the law of thermodyanamic is the restatement of the law of conservation of energy
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
