Question

How much magnesium sulfate heptahydrate is required to prepare 200 mL of 0.05 M solution? 10 g 2.46 g 0 0.05 g 12.38 24.65 g

Answer 1

To prepare a 0.05 M magnesium sulfate heptahydrate solution with a volume of 200 mL, you will need 2.46 grams of magnesium sulfate heptahydrate.

Step-by-step explanation:

To calculate the amount of magnesium sulfate heptahydrate required to prepare 200 mL of 0.05 M solution, we need to use the formula:

moles = molarity x volume

First, we convert the volume from milliliters to liters:

200 mL x (1 L/1000 mL) = 0.2 L

Next, we substitute the given values into the formula:

moles = 0.05 mol/L x 0.2 L = 0.01 mol

Finally, we calculate the mass of magnesium sulfate heptahydrate using its molar mass:

moles = mass (g) / molar mass (g/mol)

0.01 mol = mass (g) / 246.48 g/mol

mass (g) = 0.01 mol x 246.48 g/mol = 2.46 g

Answer 2

Answer: The mass of magnesium sulfate heptahydrate required is 2.46 g

Step-by-step explanation:

To calculate the mass of solute for given molarity, we use the equation:

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

We are given:

Molarity of solution = 0.05 mol/L

Molar mass of magnesium sulfate heptahydrate = 246.47 g/mol

Volume of solution = 200 mL

Putting values in above equation, we get:

`0.05mol/L=\frac{\text{Mass of }MgSO_4.7H_2O* 1000}{246.47g/mol* 200mL}\\\\\text{Mass of }MgSO_4.7H_2O=2.46g`

Hence, the mass of magnesium sulfate heptahydrate required is 2.46 g