How many moles of [tex]CH_3OH[/tex] are contained in 155 mL of a 0.167 m [tex]CH_3OH[/tex] solution? The density of the solution is 1.44 g/mL.

A. [tex]3.73 \times 10^{-2}[/tex] mol
B. [tex]1.55 \times 10^{-3}[/tex] mol
C. [tex]1.55 \times 10^{-6}[/tex] mol
D. [tex]1.34 \times 10^{-1}[/tex] mol

The molar concentration of a solution refers to the number of moles of a solute present in one litre of the solution. Therefore, it can be calculated by dividing the number of moles of solute by the volume of the solution in liters.In order to calculate the number of moles of CH3OH present in 155 mL of 0.167 mCH3OH solution, we can use the following formula:Number of moles of CH3OH = Molar concentration × Volume of solution in litersStep-by-step solution:Molar concentration of CH3OH = 0.167 m

To convert 155 mL to liters, we divide by 1000:Volume of CH3OH solution = 155/1000 L

= 0.155 LUsing the formula,

Number of moles of CH3OH = Molar concentration × Volume of solution in liters

= 0.167 mol/L × 0.155 L

= 0.025885 mol

Therefore, the number of moles of CH3OH present in 155 mL of 0.167 mCH3OH solution is 0.025885 mol (option a) 3.73×10^−2 mol).

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How many moles of [tex]CH_3OH[/tex] are contained in 155 mL of a 0.167 m [tex]CH_3OH[/tex] solution? The density of the solution is 1.44 g/mL.A. [tex]3.73 \times 10^{-2}[/tex] mol B. [tex]1.55 \times 10^{-3}[/tex] mol C. [tex]1.55 \times 10^{-6}[/tex] mol D. [tex]1.34 \times 10^{-1}[/tex] mol

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Other Questions

How many moles of [tex]CH_3OH[/tex] are contained in 155 mL of a 0.167 m [tex]CH_3OH[/tex] solution? The density of the solution is 1.44 g/mL.

A. [tex]3.73 \times 10^{-2}[/tex] mol
B. [tex]1.55 \times 10^{-3}[/tex] mol
C. [tex]1.55 \times 10^{-6}[/tex] mol
D. [tex]1.34 \times 10^{-1}[/tex] mol