im trying to do online chemistry homework. ive tried these few problems a million times and i just cant get them! help please!

calculate the number of moles of Cr2O7^2- in 31.5 mL of a .0250 M K2Cr2O7 solution.

and also

4.42g of (NH4)3PO4 were dissolved in 20.0L of water. how many moles of NH4+ does the solution contain?

one more

what is the molarity of the solution resulting from dissolving 17.4g of silver nitrate in enough water to give a final volume of 366 mL.

so for this one, i tried figuring out the mass of silver nitrate. i got 169.91. and then i think you need to use the equation moles of solute/volume of solution=molarity. but to find the moles of silver nitrate you take the 169.91 and divide it by 17.4. then take that number and divide it by 366? i dont think thats the right answer.

thanks for your help !!

Moles questions?

moles = concentration*volume

So for K2Cr2O7 moles = 0.025*.0315 (in litres)

=7.875*10^-4

As 1 mole K2Cr2O7 liberates 1 mole of Cr2O72- the number of moles of Cr2O7- = 7.875*10^-4

moles (NH4)3PO4 = mass/ MW

= 4.42/ (3*(14.0+4*1)+31.0+4*16.0) = 0.0297.

1mole of (NH4)3PO4 liberates 3 moles of NH4+

So the moles NH4+ = 0.0297 * 3 =0.0890

NOTE 20.0 ltr of water is unnecessary info

First need moles of AgNO3

= 17.4/(107.9+ 14.0+3*16.0)

= 0.1024

Conc = moles/vol in litres

= 0.1024/ 0.366

=0.280M

Edit: Mad scientist has made a mistake The MW of NH4+ is 14+1+1+1+1 = 18 not 17.

Edit 2: Problem with Mad scientist in first question. you said "5.397 g of Cr207 = 0.01835 mol". This is wrong it should be 5.397/215.9 = 0.025 (which is back to where the question started); you have divided by the MW of K2Cr2O7.

BTW: I am not having a go at mad scientist. It is just that someone posts on here to learn and get answers and they shouldn't be mislead.

Reply:right let me get passed 3rd grade and i will help u alright

BLAING!!!!!

Reply:Cr207 = 215.88 g/mol

K2Cr2O7 = 294.18 g/mol K2 = 78.3 g/mol

.025 M K2Cr2O7 = 7.3545 g (per 1000ml solution)

[Cr207 = 215.88 g/mol] / [K2Cr2O7 = 294.18 g/mol] = 73.4%

0.734 x 7.3545g = 5.397 g of Cr207 in 1000ml

5.397 g of Cr207 = 0.01835 mol

.01835mol/ 1000ml = ____mol/31.5 ml

therefore #mol = (.01835x31.5ml)/1000ml

______________________________________

NH4 = 17.04 g/mol (NH4)3PO4 = 155.12 g/mol

[(NH4)3 = 51.12 g/mol] / [(NH4)3PO4 = 155.12 g/mol] = 0.32955

0.32955 x 4.42 g = 1.4566 g NH4 / 17.04 g /mol = 0.0855 mol

______________________________________...

AgNO3 = 169.91 g/mol

17.4 g AgNO3 divided by 169.91 g/mol = 0.1024 mol

[0.1024 mol / 366ml] = [_____mol / 1000ml]

this molar amount will give you the Molarity with should be about 0.28 M

Its been a while since I've done these problems but they make sense to me.

yoga

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