What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 100. mL of solution

Answer :

Answer:

[tex]3.65~M[/tex]

Explanation:

We have to remember the molarity equation:

[tex]M=\frac{mol}{L}[/tex]

So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the conversion:

[tex]100~mL\frac{1~L}{1000~mL}=~0.1~L[/tex]

Now we can calculate the moles. For this we have to calculate the molar mass:

O: 16 g/mol

H: 1 g/mol

C: 12 g/mol

[tex](16*1)+(1*4)+(12*1)=32~g/mol[/tex]

With the molar mass value we can calculate the number of moles:

[tex]1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH[/tex]

Finally, we can calculate the molarity:

[tex]M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M[/tex]

I hope it helps!

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