All categories

3 years ago

Find the molarity and molality of 37.0 wt% Hcl. Assume the density of the acid is 1.19g/ml

Chemistry

1 answer:

erik [133]3 years ago

4 0

Answer:

a) M HCl = 32.162 mol/L

b) m HCl = 15.873 mol/Kg

Explanation:

wt% = (mass HCl / mass sln)×100
δ HCl = mass HCl / Volume sln
molarity = mol HCl / V sln (L)
molality = mol HCl / Kg solvent

∴ MW HCl = 36.46 g/mol

assuming mass sln = 1 g

∴ mass sln = 1 g = g solvent + g solute

∴ solute: HCl

⇒ 0.37 = g HCl / g sln

⇒ g HCl = 0.37 g

⇒ g ste = 1 - 0.37 = 0.63 g

⇒ Kg ste = (0.63 g)×(Kg/1000 g) = 6.3 E-4 Kg

⇒ mol HCl = (0.37 g HCl)×(mol/36.46 g HCl) = 0.010 mol HCl

∴ V sln = (0.37 g)/(1.19 g/mL) = 0.3109 mL

⇒ V sln = (0.31 mL)×( L/1000 mL) = 3.109 E-4 L

a) molarity (M):

⇒ M HCl = 0.010 mol HCl / 3.109 E-4 L

⇒ M HCl = 32.162 mol/L

b) molality (m):

⇒ m HCl = ( 0.010 mol HCl) / (6.3 E-4 Kg)

⇒ m HCl = 15.873 mol/Kg

You might be interested in

1 times 2900 inches for a car

Delicious77 [7]

One times anything is the same number. 1 x 2900= 2900

5 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

What is the source of energy for an earthquake

MA_775_DIABLO [31]

Seismic waves is the energy of and earthquake aka Elastic Energy

3 0

3 years ago

Read 2 more answers

A student mixes sugar and water until the sugar dissolves. This an example of a HELP PLS !!

murzikaleks [220]

Answer:

The sugar water mix is a mixture.

6 0

3 years ago

Find How many joules of heat energy are needed to raise the temperature of 10grams of Aluminum from 20*C to 40*C. If the specifi

dmitriy555 [2]

Explanation:

the answer and the working out is shown above, hope it helps.

3 0

2 years ago

Find the molarity and molality of 37.0 wt% Hcl. Assume the density of the acid is 1.19g/ml​

Find the molarity and molality of 37.0 wt% Hcl. Assume the density of the acid is 1.19g/ml