Answer:
Explanation:
Molarity = number of moles / volume
If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
1.34*10^6
Move the decimal 6 times to the left, it is a number 1-10
Answer:
219.15 grams
Explanation:
What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)
Mole of Na = 22.99g
Mole of Cl = 35.45g
For NaCl we have ratio of 1:1, so we have 1 Na for every Cl
So we just add the two together to get the molar mass of NaCl which is
22.99 + 35.45 = 58.44g/mol
And we know we have 3.75 moles of NaCl so we multiply that by the molar mass of NaCl to get our answer
3.75 x 58.44 = 219.15grams
