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3 years ago

How many grams of magnesium chloride are needed to make 700ml of a 1.4 m solution?

1 answer:

eduard3 years ago

4 0

Molarity is defined as the number of moles of solute in 1 L solution
molarity of MgCl₂ solution to be prepared - 1.4 M
volume of the solution to be prepared - 700 mL
the number of moles of MgCl₂ in 1 L - 1.4 M
therefore 700 mL should contain - 1.4 mol/L x 0.700 L = 0.98 mol
molar mass of MgCl₂ - 95.2 g/mol
mass of MgCl₂ required - 0.98 mol x 95.2 g/mol = 93.3 g
mass of MgCl₂ required is 93.3 g

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3 years ago

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Explanation:

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$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the new pressure, volume, and temperature. Before plugging the values into the equation, you need to

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(2) convert the temperatures from Celsius to Kelvin (°C + 273)

The final answer should have 3 sig figs like the given values.

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T₁ = 20°C + 273 = 293 K T₂ = 40°C + 273 = 313 K

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ <----- Combined Gas Law

$\frac{(0.480 atm)(5.00 L)}{293 K}=\frac{P_2(2.30 L)}{313 K}$ <----- Insert values

$0.00819=\frac{P_2(2.30 L)}{313 K}$ <----- Simplify left side

$2.56 = P_2(2.30L)$ <----- Multiply both sides by 313

$1.12 = P_2$ <----- Divide both sides by 2.30

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3 years ago

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