Question

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 1

Answer:

$3.65~M$

Explanation:

We have to remember the molarity equation:

$M=\frac{mol}{L}$

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

$100~mL\frac{1~L}{1000~mL}=~0.1~L$

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

$(16*1)+(1*4)+(12*1)=32~g/mol$

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

$1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH$

Finally, we can calculate the molarity:

$M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M$

I hope it helps!