[TheCluelessOne]

I guess we are supposed to find the Solute of the solution

0.289 liters of a 0.00300 M Cu(NO3)2 solution

16.00 mililiters of a 5.90-molar Pb(NO3)2 Solution

508 mL of a 2.75-molar NaF solution

6.20 L of a 3.76-molar Na2O solution

[Borek]

I guess you guess right.

There is a possible problem if the solution are to be prepared using hydrated salts.

[TheCluelessOne]

I dont think we are using hydrated salts

[Borek]

So, how much solute in 0.289 liters of a 0.00300 M solution?

[TheCluelessOne]

the solution i was taught was
M*Liter Solution*Molar Mass

so
0.163 grams of solute?

but the others are more complicated

[Borek]

OK so far.

There is nothing complicated in converting volume in mL to volume in L.

[TheCluelessOne]

well I don't know much about these equations     
                   
how do you convert molar to M(molality)?

[Borek]

M is usually molarity and molar means exactly the same. No need for molality here.

How many mL in 1 L?

[TheCluelessOne]

so there is no difference between M and molar?

That was the only part I was confused about

and 1000 mL in 1 L

so I just need to move a few decimal places..

[Borek]

Nothing more complicated here.

[TheCluelessOne]

so number 2 would be

31.26 grams of solute...

and number 3 would be

58.660 Grams of Solute

and number 4

1444.878 g of solute

Am I right?

[Borek]

Your sigfigs are off, but apparently you got the idea.

[TheCluelessOne]

So how do you go about Sigfigs?

make for example #2

16.00 mL of a 5.90 M Pb(NO3)2 Solution

when I convert 16.00 mL to L its 0.1600 mL and I know that 1 and 6 are sigfigs, do i have to include the 2 zero's too?

so it makes 4 sigfigs?