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part iv co convert volume and molarity to mass [hint: first convert ml to l] 8. what mass of potassium iodide (ki) is needed to prepare

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Part IV Co Convert volume and Molarity to mass [Hint: First convert mL to L] 8. What mass of potassium iodide (KI) is needed to prepare 340 mL of a 2.8 M solution? 9. What mass of sodium chloride (NaCl) is required to prepare 635 mL of a 6.0 M solution? 10. What mass of magnesium citrate (MgC_(6)H_(6)O_(7)) is required to prepare 115 mL of a 070 M solution? The molar mass of magnesium citrate is 214.41g/mol

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Answer

## For question 8:The volume is 340 mL, which is L. The number of moles is moles. The mass of KI needed is g. ## For question 9:The volume is 635 mL, which is L. The number of moles is moles. The mass of NaCl needed is g. ## For question 10:The volume is 115 mL, which is L. The number of moles is moles. The mass of needed is g.

Explanation

## Step1:We'll use the concept of molarity (M) in chemistry. Molarity is calculated as the number of moles of solute per liter of solution. ## Step2:We'll use the formula for molarity , where is the number of moles and is the volume in liters. ## Step3:We'll rearrange the formula to solve for the number of moles . ## Step4:We'll convert the volume from milliliters to liters by dividing by 1000. ## Step5:We'll substitute the given molarity and volume into the formula to find the number of moles. ## Step6:We'll then multiply the number of moles by the molar mass to find the mass of the solute needed. ### For question 8: The molar mass of potassium iodide (KI) is 166 g/mol. ### For question 9: The molar mass of sodium chloride (NaCl) is 58.44 g/mol. ### For question 10: The molar mass of magnesium citrate \((MgC_{6}H_{6}O_{7})\) is 214.41 g/mol.