What mass of solute is required to prepare a #0.875*"molal"# solution from #534*mL# of water solvent?

Question

Noah Aldrich · Accepted Answer

Approx. #27*g#. I assume you mean a #"molal concentration"#.

#"Molality"="moles of solute"/"kg of solvent"# Thus #0.875*mol*kg^-1# #=# #"number of moles of NaCl"/(0.534*kg)# And #"number of moles of NaCl"# #=# #0.875*mol*kg^-1xx0.534*kg# #=# #0.467*mol# #"Grams of NaCl"=0.467*molxx58.44*g*mol^-1#. The #"molar concentration"# would be to all intents and purposes identical.

Natalie Anton · Answer

undefined To calculate the mass of solute required to prepare a 0.875 molal solution from 534 mL of water solvent, you can use the formula:

Mass of solute (in grams) = molality (in mol/kg) × molecular weight of solvent (in g/mol) × mass of solvent (in kg)

First, convert the volume of water solvent from milliliters to kilograms:

534 mL × (1 g / 1 mL) × (1 kg / 1000 g) = 0.534 kg

Then, multiply the molality by the molecular weight of the solvent (which is approximately 18.015 g/mol for water) and the mass of the solvent:

0.875 mol/kg × 18.015 g/mol × 0.534 kg = 7.18 grams (rounded to two decimal places)

Therefore, approximately 7.18 grams of solute is required to prepare a 0.875 molal solution in 534 mL of water solvent.