The Student Room Group

Reply 1

jacquelyn
i really do not understand how you do these type of mole questions:

what is the concentration in moles dm-3 of the following?

21. 3.65g of HCL in 1000cm3 of solution?

31. 240cm3 of NH3 (g) dissolved in 1000cm3 of solution?


21. Moles of HCl = mass/Mr = 3.65/36.5 = 0.1 Moles
Concentration of HCL = Moles/Volume(in dm^-3) = 0.1/1.000 = 0.1 moles dm-3

31. Moles of NH3 = 240/24000 = 0.01 Moles (assuming it was dissolved into solution at room temp)
Concentration of NH3 = Moles/Volume = 0.01/1.000 = 0.01 moles dm-3

Reply 2

jacquelyn
i really do not understand how you do these type of mole questions:

what is the concentration in moles dm-3 of the following?

21. 3.65g of HCL in 1000cm3 of solution?

31. 240cm3 of NH3 (g) dissolved in 1000cm3 of solution?

A 1M solution (1 mol dm^-3) is defined to have 1 mole of compound dissolved in 1 cubic dm of water (hence 1 mol dm^-3, or 1 mole per cubic dm). So you can use the equation:

Moles = molarity (in mol dm^-3) x volume (in dm^3)

You can see from that that the dm^3 cancel on the RHS, giving you a quantity in moles on the RHS. Just manipulate that as you need, but don't forget that your molarities must have the same volume component as your 'volume' term (so change the units of the volume accordingly by multiplying or dividing by cubic scale factors).

Ben

Reply 3

thanks very much but when i tried it with 6.62g of pb(no3)2 in 250cm3 of solution it wouldn't work out

Reply 4

jacquelyn
thanks very much but when i tried it with 6.62g of pb(no3)2 in 250cm3 of solution it wouldn't work out

6.62/269 = 0.0246

0.0246 = M x (250/1000)

M = 0.0984 mol dm^-3

Ben

Reply 5

Isn't the Molar Mass of Pb(NO3)2..331...(207+[14*2]+[16*6]) ?...

If you replace 269 with 331 in Ben's reply, you get the right answer...I think. :confused:

..I agree with Ben's method...it's probably the easiest... :smile:

Reply 6

fisfos815
Isn't the Molar Mass of Pb(NO3)2..331...(207+[14*2]+[16*6]) ?...

If you replace 269 with 331 in Ben's reply, you get the right answer...I think. :confused:

..I agree with Ben's method...it's probably the easiest... :smile:

Yes, it is - I must've only included one of the NO3 groups.

Sorry about that!

Ben