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3 years ago

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In an experiment, a student dissolves 56.3g of MgCl2 in 0.12kg of solvent. What is the molality of this solution? Show your work

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1 answer:

Kazeer [188]3 years ago

7 0

<u>Answer:</u> The molality of magnesium chloride is 4.93 m

<u>Explanation:</u>

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

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

Where,

$m_{solute}$ = Given mass of solute $(MgCl_2)$ = 56.3 g

$M_{solute}$ = Molar mass of solute $(MgCl_2)$ = 95.2 g/mol

$W_{solvent}$ = Mass of solvent = 0.12 kg

Putting values in above equation, we get:

$\text{Molality of }MgCl_2=\frac{56.3}{95.2\times 0.12}\\\\\text{Molality of }MgCl_2=4.93m$

Hence, the molality of magnesium chloride is 4.93 m

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3 years ago

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6NaBr+1AlO3=3Na2O+2AlBr3 How many grams of NaBr would be needed in order to make 23.5 grams of AlBr3

Evgesh-ka [11]

Answer:

23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr

Explanation:

The balanced equation here is

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3 years ago

Using a spectrophotometer, and a cuvette with a path length of 1 cm you measure the absorbance (A275) of Guanosine to be 0.70. C

Anton [14]

Answer : The concentration of guanosine in your sample is, $8.33\times 10^{-5}M$

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution = 0.70

C = concentration of solution = ?

l = path length = 1.00 cm

$\epsilon$ = molar absorptivity coefficient guanosine = $8400M^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$0.70=8400M^{-1}cm^{-1}\times C\times 1.00cm$

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3 years ago

A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an

inysia [295]

<u>Answer:</u> The molar mass of the unknown compound is 223.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 8.44 torr

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $62.3637\text{ L torr }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $4.54\times 10^{-4}M$

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Volume of solution = 150.0 mL

Putting values in above equation, we get:

$4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol$

Hence, the molar mass of the unknown compound is 223.2 g/mol

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