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Answer:

A. 60 kPa

Explanation:

P2 = p1 times t2 / t1

Answer:

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

Learn more about percentage mass:

brainly.com/question/3940233

Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by  

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of  

hydration must be included.

1 Manganes  52.94 g = 63.55 g  

1 Sulphur  32.07 g =  

32.07 g 2 Hydrogen is  = 2.02 g

4 Oygen       =  

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol  18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

Answer:

The freezing point of a solution is lowered compared to the freezing point of the pure solvent. The amount of depression of the freezing point is proportional to the molality of the solute. The greater the molality of a solution, the lower its freezing point. To compare the freezing points of these solutions, we need to determine which one has the highest molality.

First, we need to determine the number of particles that each solute will produce in solution, as this affects the amount of depression of the freezing point.

KNO3 dissociates into two ions: K+ and NO3-, so it will produce two particles per formula unit.

BaCl2 dissociates into three ions: Ba2+ and two Cl-, so it will produce three particles per formula unit.

Ethylene glycol does not dissociate in solution, so it will produce one particle per molecule.

Na3PO4 dissociates into four ions: three Na+ and one PO43-, so it will produce four particles per formula unit.

Now, we can calculate the molality (moles of solute per kilogram of solvent) for each solution:

For 0.10 m KNO3: molality = 0.10 mol / 1 kg = 0.10 m

For 0.10 m BaCl2: molality = 0.10 mol x 3 particles / 1 kg = 0.30 m

For 0.10 m ethylene glycol: molality = 0.10 mol / 1 kg = 0.10 m

For 0.10 m Na3PO4: molality = 0.10 mol x 4 particles / 1 kg = 0.40 m

So, the solutions in order of decreasing freezing points are:

0.10 m Na3PO4 (highest molality)

0.10 m BaCl2

0.10 m KNO3 and 0.10 m ethylene glycol (same molality, but KNO3 has a smaller van't Hoff factor than ethylene glycol, so it will have a slightly higher freezing point)

Explanation: