How many grams of solute are needed to prepare 2 L of a 1 M solution? (Molar mas

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Question: How many grams of solute are needed to prepare 2 L of a 1 M solution? (Molar mass = 58.5 g/mol)

Options:

Correct Answer: 117 g

Solution:

Grams of solute = moles x molar mass = 2 moles x 58.5 g/mol = 117 g.

How many grams of solute are needed to prepare 2 L of a 1 M solution? (Molar mass = 58.5 g/mol)

How many grams of solute are needed to prepare 2 L of a 1 M solution? (Molar mass = 58.5 g/mol)

Step 1: Understand that a 1 M (molar) solution means there is 1 mole of solute in 1 liter of solution.
Step 2: Since we need to prepare 2 liters of a 1 M solution, we need 2 moles of solute (1 mole/L x 2 L = 2 moles).
Step 3: Know the molar mass of the solute, which is given as 58.5 g/mol.
Step 4: Calculate the grams of solute needed by multiplying the number of moles by the molar mass: 2 moles x 58.5 g/mol.
Step 5: Perform the multiplication: 2 x 58.5 = 117 grams.

Molarity – Molarity (M) is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution.
Moles and Molar Mass – To find the mass of a solute, you multiply the number of moles by the molar mass of the solute.