What quantity, in grams, of methanol, ch3oh, is required to prepare a 0.244 m solution in 400. g of water?

3.1213 g of methanol will required to  prepare 0.244 m of solution in 400 g of water.

Explanation;

Let the mass of methanol be x

Mass of the solvent = 400 g= 0,4 kg

1 kg = 1000 g

Molality of solution = 0.244 m

Molality=

3.1213 g of methanol will required to  prepare 0.244 m of solution in 400 g of water.

it's all four options.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.