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2 years ago

A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19 g/mL at 25°C

Chemistry

2 answers:

Nadya [2.5K]2 years ago

7 0

1. A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19 g/mL at 25°C ---What is the Molarity?

Molarity = n/v

Where n = moles and v = liters of solution

density = mass / volume (for the solution)

find mass of solution

m = d*v = 1.19 * 1000 = 1190 g solution

Next, lets calculate the mass of the NaHCO₃:

7.06% means 7.06 g of NaHCO₃ ---------> 100 g of NaHCO₃

we want to know how many "g" of NaHCO₃ in a 1190 g solution.

7.06 * (1190/100) = 84.04 g NaHCO₃

Next, we want to find the number of moles of NaHCO₃:

n = mass NaHCO₃ / Molar Mass = 84 (rounded) / 84 = 1 mol

Answer for question one = 1 Mol

2. A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19 g/mL at 25°C ---- What is the Molality?

Molality = n / 1 kg = n / mass H₂O

mass H₂O = 1190g solution - 84.01 g NaHCO₃ = 1100 g H₂O = 1.106 kg H₂O

Molality = 1 mol NaHCO₃ / 1.106 kg H₂O = 0.904 mol / kg

Molality = 0.904 mol / kg

This is as far as I can get, since I am in middle school. The rest is too difficult and I have a massive packet for physics to complete. Sorry I can't help further than this. I hope it helps a little though. Also, sorry because my answer is unorganized.

Ronch [10]2 years ago

4 0

Q1:

Answer:

M = 1.0 mol/L.

Explanation:

• To solve this problem, we can use the relation:

M = 10Pd / Molar mass, where

M is the molarity of the solution (mol/L),

P is the percent of the solution (P = 7.06 %),

d is the density of the solution (d = 1.19 g/ml),

Molar mass of sodium bicarbonate (84.007 g/mol),

M = 10 (7.06 %) (1.19 g/ml) / (84.007 g/mol) = 1.0 mol/L.

Q2:

Answer:

Molality = 0.904 m.

Explanation:

• To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

The solute is sodium bicarbonate (7.06 %), it is the component with the lower percent.

The solvent is water (92.94 %), it is the component with the higher percent.

• To get the mass of the solute and solvent, we should firstly get the mass of the solution.

Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (1.19 g/ml) (1000 ml) = 1190 g.

The mass of solute sodium bicarbonate = (the total mass of the solution x percent % of solute) / 100 = (1190 x 7.06) / 100 = 84.014 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (1190 x 92.94) / 100 = 1105.986 g.

m = (mass / molar mass) solute x (1000 / mass of solvent)

m = (84.014 g / 84.007 g/mol) (1000 / 1105.986 g) = 0.904 m.

Q3:

Answer:

Mole fraction of solvent (water) = 0.953.

Explanation:

• The solute is methanol CH3OH (8.05 %), it is the component with the lower percent.

• The solvent is water (91.95 %), it is the component with the higher percent.

• To solve this problem, we can use the relation of mole fraction of water (X water):

X water = (no. of moles of water) / (total no. of moles of the solution)

• To get the no. of moles of the solute and solvent, we should obtain the mass of both.

• To obtain the mass of the solute and solvent, we should firstly get the mass of the solution.

• Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (0.976 g/ml) (1000 ml) = 976.0 g.

The mass of solute methanol CH3OH = (the total mass of the solution x percent % of solute) / 100 = (976.0 x 8.05) / 100 = 78.568 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (976.0 x 91.95) / 100 = 897.432 g.

The no. of moles of solute methanol CH3OH = mass / molar mass = (78.568 g) / (32.0 g/mol) = 2.45525 mol.

The no. of moles of solvent water = mass / molar mass = (897.432 g) / (18.0 g/mol) = 49.857 mol.

X water = (no. of moles of water) / (total no. of moles of the solution) = (49.857) / (49.857 + 2.45525) = 0.953.

Q4:

Answer:

m = 5.18 m.

Explanation:

• To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

Mass of the solvent water = 60.0 g.

m = (mass / molar mass) cesium chloride x (1000 / mass of water)

m = (52.3 g / 168.36 g/mol) (1000 / 60.0 g) = 5.177 m = 5.18 m.

Q5:

Answer:

4.91 M.

Explanation:

• To solve this problem, we can use the relation of molarity (M):

M = (mass / molar mass) solute x (1000 / V of the solution)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

V of the solution = 63.3 ml.

M = (mass / molar mass) cesium chloride x (1000 / V of the solution)

M = (52.3 g / 168.36 g/mol) (1000 / 63.3 g) = 4.907 M = 4.91 M.

Q6:

Answer:

The vapor pressure of the solution = 80.0 torr.

Explanation:

• The vapor pressure of the solution will be the partial vapor pressure of the ethanol only because alpha naphthol is assumed to be nonvolatile, so it has no partial vapor pressure in the total vapor pressure of the solution.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol),

Where, X ethanol is the mole fraction of ethanol.

P0 ethanol is the vapor pressure of pure ethanol (P0 = 100.0 torr).

• To get the mole fraction of ethanol, we can use the relation of mole fraction of ethanol (X ethanol):

X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution).

• The no. of moles of ethanol = mass / molar mass = (36.8 g) / (46.07 g/mol) = 0.80 mol.

• The no. of moles of alpha naphthol = mass / molar mass = (28.8 g) / (144.0 g/mol) = 0.2 mol.

• X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution) = (0.80) / (0.80 + 0.2) = 0.80.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol) = (0.80) (100.0 torr) = 80.0 torr.

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