chemistry wizard plz help?

Question

How much solute is needed to make 1.49 L of
a 3.42 M solution of Ba(NO3)2? Answer in
units of g.

Answer 1

Molarity is moles solute / liters solution.

3.42 mol / 1 L = x mol / 1.49 L

x = moles needed

Once you calculte moles needed you multiply that by the molar mass of Ba(NO3)2

Answer 2

moles needed = molarity * volume = 3.42 * 1.49 = 5.10 moles

MW Ba(NO3)2 = 137.3 + 2(14) + 6(16) = 261.3 g/mol

mass = moles * molar mass = 5.10 * 261.3 = 1332.63 g

Answer 3

molarity = moles/liter

So, 3.42 = moles/1.49L

moles = 5.095 moles

Now, moles = mass/molecular weight

plug the value of moles and molecular weight in above euqation and solve for the mass (g).

5.095 = mass/261

mass = 1329.795 g