Question

A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

Answer 1

We need 0.375 mol of CH3OH to prepare the solution

Step-by-step explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

`d = (g)/(mL) \\\\ V= (g)/(d) = (500g)/(1 (g)/(mL) ) = 500mL = 0,5 L`

Now, find moles of
`CH_(3) OH` are needed using the molarity equation:

`M = ( moles )/( V (L)) \\\\\\molesCH_(3)OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_(3)OH = 0,375 mol`

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M