Chem Question Find Molarity

Chem Molarity Question for FeSO4 and K2Cr2O7, finding molarity of Cr+3?

moles Fe2+ = 0.0500 L x 0.761 M = 0.0381
moles Cr2O72- = 0.0200 L x 1.19 M = 0.0238

Cr2O72- + 6 Fe2+ + 14H+ >> 2 Cr3+ + 6Fe3+ + 7H2o

1 : 6 = x : 0.0381
x =0.00635 = moles Cr2O72- that react with Fe2+

this will give 2 x 0.00635 = 0.0127 moles Cr3+

total volume = 0.0700 L
[Cr3+]= 0.0127/ 0.0700= 0.181 M

Molarity question for chem lab?

You have to express the molarity of the SCN- ion. Look at the part of the question that says:
2ml of 0.002M NaSCN
NaSCN dissociates:
NaSCN = Na + + SCN-
The concentration of the SCN- ion is the same as the concentartion of the NaSCN. That is it is 0.002M

You had 2ml of this solution. You add 10ml of Fe(NO3)2 and 8ml of HNO3. With the original 2ml of NaSCN solution, you now have a final volume of 20ml

What is the new concentration of SCN- ion?

M1V1 = M2V2
M1*20 = 0.002*2
M1 = 0.004*20
M1 = 0.0002M

The concentration of the SCN- ion in the final solution = 0.0002M

Molarity Chemistry Question?

anytime you see a question about "how much of this or that reacts with or is produced" or "what is the yield" or "what is the % yield" or even... like in this case.. "what is the concentration of a reaction".. you should immediately think..

"this is a stoichiometry problem"

and.. the steps to solving stoichiometry problems are... .(MEMORIZE THESE)
.. (1) write a balanced equation
.. (2) convert everything to moles
.. (3) determine limiting reagent
.. (4) convert moles limiting reagent to moles other chemical species
.. (5) convert moles back to mass.. this is theoretical mass (yield)
.. (6) % yield = actual mass recovered / theoretical mass x 100%

the idea being the coefficients of balanced equations are in MOLE ratios. So we convert to moles first, then use those coefficients to convert between different chemicals.... then back to mass again or whatever units you need.

here we go
*** 1 ***
1 Cu(2+)(aq) + 1 S(2-)(aq) --> 1 CuS(s)

note the ratios.. 1:1 Cu2+ to S2-

*** 2 ***
moles S(2-)
.. 30.8mL x (1L / 1000mL) x (0.267 mol S(2-) / L) = 0.00822

*** 3 ***
S(2-) is the limiting reagent for this problem. We know that value and we're free to calculate moles of Cu(2+)

*** 4 ***
0.00822 mol S(2-) x (1 mol Cu(2+) / 1 mol S(2-) = 0.00822 mol Cu(2+)
..... .... .... .... .... .... ...↑... .... .... .... ...↑
.... ... .. coefficients of the balance equation
... this converts moles S(2-) to moles Cu(2+)

*** 5 ***
now for this problem, we're not looking for mass, we're looking for molarity.. moles Cu(2+) per liter of solution
... M = 0.00822mol Cu(2+) / 0.0356L = 0.231M

so the last choice is the correct one.

********
********
now.. you should be using "dimensional analysis" and therefore can combine all those steps into 1. And you can molarity as moles / L.... AND... as millimoles / milliliter... mmol / mL

so that..
.. 30.8mL S(2-) x (0.267 mmol S(2-) / mL S(2-) x (1 mmol Cu(2+) / mmol S(2-)
... . = 8.22mmol Cu(2+)
and then
... M = 8.22 mmol Cu(2+) / 35.6mL = 0.231M

that's how I would do it... notice that I've combined steps 2 and 4 into 1 equation?

Help with a chem molarity question?

Take 1 litre (1000 ml) of the NaCl solutio
Because density = 1.210g/ml, mass = 1210g
NaCl = 10.9% = 10.9/100*1210 = 131.89g

Molar mass NaCl = 22.99+35.453 = 58.443g/mol
131.89g = 131.89/58.443 = 2.256 mol

Molarity of solution = 2.256M

Caculate molality:
In the 1210g of solution you have 131.89g NaCl
Therefore mass of water = 1210-131.89 = 1078.11g
Mass of NaCl dissolved in 1000g water = 131.89/1078.11*1000 = 122.3
122.3g NaCl = 2.09moles
Molality of solution = 2.09m

NOTE: Ashish Sinha and I both get the same answers from different methods of solving. But there is a serious problem with this question. Look at the quantity of water used as solvent in my calculation of molality: The figures tell us that a solution of NaCl with a volume of 1000ml contains 1078.11 g or ml of solvent, which is impossible. Where is the problem? This lies with the stated density of the solution at 1.210g/ml. This is impossible. If you add 109g of NaCl to 1000g water, the total mass will be 1109g, with density = <1.109g/ml ( a rough calculation) My reference quotes the density of 10.9% NaCl solution as 1.077g/ml. Working with the incorrect density has produced totally incorrect figures. But the mathematics of the two solutions at least are correct, which is what you want to know.

Analytical Chemistry: Molarity Question?

First, yu need to calculate the molecular weight of glocose. You can either look it up on line or calculate it from the formula you have given above. C = 6 x 12, H = 1 x 12, O = 6 x16. Add those up and you get 180 g/mole.

So, I'll work out the first part and you can do the last part:

(80 mg/100 mL) /180g/mole = 0.444 mole/ 100 L = 0.00444 moles/L - 0.00444 M

(the milli in grams and the milli in liters cancel.)

So just substitute the 120 mg for the 80 in the equation above and solve.

Chemistry question involving molality and molarity.?

Molarity of Water :
Molecular Wt of water =18;
Density of water at 25 C =0.997 g/cc
Wt. of 1000 cc of water =997 gms

Hence Molarity = 997/18 =55.39 M;

Molality of Water :
Here we have to see the Gm moles in 1000gms of solvent ;

Hence Molality =1000/18 =55.55 M

Chem help find the molarity?

M HCl = (0.03312 L HCl/0.1 L sol'n) (1168.4 g HCl/ L HCl) (1 mol HCl/ 37.47 g HCl) = 1.033 M

How to find molarity in a percent solution?

A 10% solution is one where the mass of the solute dissolved in the solution is equal to 10% of the volume of the solution

So 10% w/v C6H12O6 = 10 g / 100 ml of solution

Molarity = moles / litres
moles = mass / molar mass

molar mass C6H12O6 = (6 x 12.0) + (12 x 1.0) + (6 x 16.0) = 180.0 g/l
therefore moles sucrose = 10g / 180.0 g/mol
= 0.05556 moles of C6H12O6

This is in 100 ml (0.10 L) of solution

So M = moles / L
= 0.05556 mol / 0.10 L
= 0.556 M

Chemistry: Calculating Molarity?

Molarity refers to concentration expressed as moles/L.

In this problem you have a solution 0.900 M of Na3PO4

In order to know the molarity of each individual ion you must use stechiometry.

There are 3 ions of Na and 1 ion of PO4 in each ionic compound of Na3PO4.

So,

? M Na = 0.900 moles/L Na3PO4 (3 moles Na/ 1 mol Na3PO4) = 2.7 moles/L Na

? M PO4 = 0.900 moles/L Na3PO4 (1 mol PO4/1 mol Na3PO4) = 0.900 moles/L PO4