Answer:
c. 77 %
Explanation:
Percent mass (% mass) of solute = mass of solute/mass of solution × 100
According to this question, a mountain dew solution weighing 300grams contains 231 g of sugar. This means that:
% mass of sugar = 231g/300g × 100
% mass of sugar = 0.77 × 100
% mass of sugar = 77%.
Answer:
3.4752 moles of water
Explanation:
There are 13.84 mole in one cup of water so,
13.84 divided by 4= 3.4725 :)
The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)

There are 158.4 grams of CO2 in 3.6 mol of CO2.
<h3>HOW TO CALCULATE MASS?</h3>
The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass. That is;
mass of CO2 = no. of moles × molar mass
According to this question, there are 3.6 moles of CO2.
mass of CO2 = 3.6 moles × 44g/mol
mass of CO2 = 158.4g.
Therefore, there are 158.4 grams of CO2 in 3.6 mol of CO2
Learn more about mass at: brainly.com/question/15959704
Answer:
0.774g of ethanol
0.970mL of ethanol
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).
14.0g of butanol are <em>0.014kg </em>and as you want to prepare the 1.2m solution, you need to add:
0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol
To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:
0.0168moles Ethanol ₓ (46.07g / mol) =
<h3>0.774g of ethanol</h3>
And to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):
0.774g ₓ (1mL / 0.798g) =
<h3>0.970mL of ehtanol</h3>