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

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Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Determine the volume in liters**: Convert the given volume from milliliters to liters. The volume is $(200 + Y) \, ext{ml}$, which is equivalent to $\frac{200 + Y}{1000} \, ext{L}$. ‖ ‖ 2 ⋮ **Calculate the moles of solute required**: Use the molarity formula $M = \frac{n}{V}$, where $M$ is the molarity, $n$ is the number of moles, and $V$ is the volume in liters. For a $2.8 \, ext{M}$ solution, the moles of solute required are $n = M imes V = 2.8 imes \frac{200 + Y}{1000}$. ‖ ‖ 3 ⋮ **Calculate the mass of Methanol (CH$_3$OH)**: The molar mass of Methanol is $32.04 \, ext{g/mol}$. The mass required is $ ext{mass} = ext{moles} imes ext{molar mass} = 2.8 imes \frac{200 + Y}{1000} imes 32.04$. ‖ ‖ 4 ⋮ **Calculate the mass of Sodium Carbonate (Na$_2$CO$_3$)**: The molar mass of Sodium Carbonate is $105.99 \, ext{g/mol}$. The mass required is $ ext{mass} = ext{moles} imes ext{molar mass} = 2.8 imes \frac{200 + Y}{1000} imes 105.99$. ‖ ∻Answer∻ ⚹ The mass of Methanol required is $2.8 imes \frac{200 + Y}{1000} imes 32.04 \, ext{g}$. ⚹ ⚹ The mass of Sodium Carbonate required is $2.8 imes \frac{200 + Y}{1000} imes 105.99 \, ext{g}$. ⚹ ∻Key Concept∻ ⚹ Molarity and mass calculation for solution preparation ⚹ ∻Explanation∻ ⚹ To prepare a solution of a specific molarity, you need to calculate the number of moles of each solute required and then convert that to mass using the molar mass of each solute. ⚹◊What is the formula to calculate the number of grams required to prepare a solution of a given molarity and volume (in terms of moles)?⍭ Generate me a similar question◊

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