Empirical formula of c3h12o6
1 answer:
Answer:
Explanation:
The empirical formula is the simplest whole-number ratio of the elements in a compound to one another. To find the empirical formula given a molecular formula, divide all the subscripts by their greatest common factor.
3, 12, and 6 can all be divided by 3, which makes the empirical formula
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a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:
First, we have to assume that we have 100 grams of solution. This will simplify the calculations.
Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:
To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:
So, the mole fraction of NaOH is 0,073
b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:
First, we have to assume that we have 100 grams of solution. This will simplify the calculations.
Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:
Now, we apply the definition of molality to calculate the molality of the solution:
So, the molality of this solution is 4,41 m
Have a nice day!
a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:
First, we have to assume that we have 100 grams of solution. This will simplify the calculations.
Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:
To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:
So, the mole fraction of NaOH is 0,073
b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:
First, we have to assume that we have 100 grams of solution. This will simplify the calculations.
Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:
Now, we apply the definition of molality to calculate the molality of the solution:
So, the molality of this solution is 4,41 m
Have a nice day!