7) For preparation of $(200+\mathrm{Y}) \mathrm{ml}, 2.8 \mathrm{M}$ solution from the following solutes, how many grams is required?
I. Methanol
II. Sodium Carbonate

Question

7) For preparation of $(200+\mathrm{Y}) \mathrm{ml}, 2.8 \mathrm{M}$ solution from the following solutes, how many grams is required?
I. Methanol
II. Sodium Carbonate

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the amount of substance needed**: To prepare a $2.8 \mathrm{M}$ solution, we need to calculate the moles of solute required for the given volume. The molarity (M) is defined as moles of solute per liter of solution. ‖ ‖ 2 ⋮ **Methanol (CH3OH) calculation**: The molar mass of methanol is $32.04 \mathrm{g/mol}$. Using the formula $n = M imes V$, where $n$ is the number of moles, $M$ is the molarity, and $V$ is the volume in liters, we can calculate the number of moles of methanol required. ‖ ‖ 3 ⋮ **Sodium Carbonate (Na2CO3) calculation**: The molar mass of sodium carbonate is $105.99 \mathrm{g/mol}$. Similarly, we use the formula $n = M imes V$ to calculate the number of moles of sodium carbonate required. ‖ ‖ 4 ⋮ **Convert moles to grams**: To find the mass in grams, we multiply the number of moles by the molar mass of the solute. ‖ ∻Answer∻ ⚹ For Methanol (CH3OH): [Insert final answer for methanol here] ⚹ ⚹ For Sodium Carbonate (Na2CO3): [Insert final answer for sodium carbonate here] ⚹ ∻Key Concept∻ ⚹ Molarity is used to calculate the amount of solute needed to prepare a solution of a certain concentration. ⚹ ∻Explanation∻ ⚹ By knowing the molarity and the volume of the solution, we can calculate the number of moles of solute required. Multiplying the number of moles by the molar mass of the solute gives us the mass in grams needed to prepare the solution. ⚹ Please note that the final answers for methanol and sodium carbonate are not provided here as the volume (200+Y) ml is not specified with a numerical value for Y. To complete the calculation, Y must be given a specific value.◊How can we calculate the amount of grams required to prepare a $2.8 \, ext{M}$ solution of Methanol and Sodium Carbonate in a $200+ ext{Y} \, ext{ml}$ solution?⍭ Generate me a similar question◊

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