Hello, A. K.,
It's not clear which aspect of these questions you'd like help on. Please add what you've already done,or know how to do, so that the responses can be more helpful. I'll work through the first two questions with enough detail to provide guidance for the last.
What mass of (C12H22O11) is required to prepare 4.6L of a 0.4M solution?
M, or Molar, is a unit of concentration that means moles/liter. We want a 0.4M solution, so we need 0.4 moles of C12H22O11 per every liter. Add the individual atomic masses for each atom in the compound to find the molecular weight, or molar mass. 12 C atoms at 12 amu each is 144 amu, added to 22 hydrogens at 1 each, on so on. I find a total molecular weight of this compound of 342.1 g/mole. 342.1 grams will provide 1 mole of C12H22O11.
We want 4.6 liters of a 0.4M solution. (4.6 liters)*(0.40 moles/liter) = 1.84 moles
(1.84 moles)*(342.1 g/mole) = 629.5 grams C12H22O11 added to 4.6 liters of solvent, generally water.
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What volume of a 0.7M solution can be prepared with 45g of copper (I) nitrate (CuNO3)?
As before, we need the molar mass of the compound in question, now CuNO3. I find 125.55 g/mole. 45 grams CuNO3 is
45 grams CuNO3/125.55 g/mole CuNO3 = 0.358 moles of CuNO3
Wow. Less than a mole and we want a 0.7M solution? Well, here goes:
Using all the CuNO3, we can find the volume of water, X, needed to make it 0.7M by:
(0.358 moles CuNO3)/X = 0.7 moles/liter
X = 0.511 liters, or 511 ml.
We can check this by: 0.358 moles CuNO3/0.511 liters = 0.70 M CuNO3, ao we are good.
What volume of a 1.1M solution can be prepared with 100g of Mg(NO3) 2)
Mg(NO3)2 has a molar mass of 148.5 grams/mole.
Find the number of moles by dividing 100g by the molar mass and use the same approach as in the last problem to find volume.
I hope these help. Ask questions if there is a point that needs clarification.
Bob
