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sesenic [268]
2 years ago
11

How do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved

Chemistry
1 answer:
IRINA_888 [86]2 years ago
3 0

Answer:

Molar mass of MgCl2 is 95 g/mol

Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol

moles = mass given/ molar mass

= 2.7/95 = 0.028 mol

volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)

molarity of MgCl2 = moles/volume

= 0.028/0.25

= 0.112 mol/dm3

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Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

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