Calculate the volume in millimeters..?

can you please show me step by step on how to get the answer?

12.5 g Na2CO3 from a 0.120 M solution

0.850 mol NaNO3 from a 0.500 M solution

30.0 g LiOH from a 2.70 M solution

Answer:
12.5 g Na2CO3 is 12.5 g / 106 g per mole = 0.1179 moles
Liters x Molarity = moles
Liters x 0.120 = 0.1179
Liters = 0.983 L

0.850 mole NaNO3 from a 0.500 M solution
Liters x Molarity = moles
Liters x 0.500 = 0.850 moles
Liters = 1.700

30.0 g LiOH from a 2.70 Molar solution
30.0 g / 23.9 g per mole = 1.255 moles
Liters x Molarity = moles
Liters x 2.70 = 1.255
Liters = 0.465
First determine the molecular weight of each of the compunds:

Na2CO3 has 2 sodiums, one carbon, and 3 oxygens, therefore MW= 2(23) + 1(12) + 3(16) = 106

by definition a 1 M solution of Na2CO3 contains 106 grams of the compound per Liter. Therefore a 0.120 M soln. contains 0.120(106) g/L or 12.72 g/L

you want 12.5 g so divide 12.5 by 12.72 = .983
That is how many liters of 0.120 M Na2CO3 you need to get 12.5 g of Na2CO3.

Do the same for the others =)

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