<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
(4.184 J/g·°C) x (1000 g) x (28.0 - 12.0)°C = 66944 J required
(66944 J) / ((333 J/g) + ((4.184 J/g·°C) x (12.0 - 0)°C)) = 175 g ice
Answer:
The velocity of an object is the rate of change of its position with respect to a frame of reference.
Explanation:
The concentration of the final solution will be 0.2886 M.
<h3>Molar concentration</h3>
50.6 g of NaCl is dissolved in 600 mL of water.
Mole of 50.6 g NaCl = 50.6/58.44 = 0.8658 mol
Molar concentration of the oiriginal solution = 0.8658/06 = 1.4431 M
25 mL of 1.4431 M of the solution is diluted up to the 125 mL mark.
m1 = 1.4431, v1 = 25 ml, v2 = 125 mL
m2 = m1v1/v2 = 1.4431 x 25/125 = 0.2886 M
The concentration of the final solution will be 0.2886 M
More on molar concentration can be found here: brainly.com/question/15532279
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Answer:
35.94 g of NaCl
Explanation:
First write a balanced chemical reaction.
2Na + Cl₂ --> 2NaCl
(Note that the Cl has a subscript of 2 because it's diatomic)
Then, set up an basic stoichiometry equation using one of the given masses of the reactants. I chose chloride.
The masses were acquired from the periodic table. (compound's mass requires the addition of individual element's masses) Also, the number of moles per reactant or product is based on the coefficient from the chemical reaction. I can't show this, but note that the units do cancel out so all that's left is grams of sodium chloride, which is what we were looking for.