A Solution that contains 53.6 g of an unknown compound in 241.2 g of water freez

Question

A Solution that contains 53.6 g of an unknown compound in 241.2 g of water freezes at -8.00 degree Calculate the molar mass (in units of g/mol) of the solute. K_f of water is 1.86 degree C/m. 8.69E-1 g/mol 1.49E1 g/mol 1.05E3 g/mol 3.01E3 g/mol 5.17E1 g/mol What is the molarity of a solution of 30.% by mass cadmium sulfate. CdSO_4 (molar mass = 208.46 g/mol) by mass? The density of the solution is 1.10 g/mL. 1.31E0 M 1.44E0 M 1.58E1 M 1.58E-2 M 1.58E0 M 0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20 degree Calculate the molar mass of the compound. 663 g/mol 1.10 times 10^2 g/mol 727 g/mol 1.15 g/mol 0.872 g/mol The reaction A + 2B rightarrow products was found to follow the rate law: rate = k[A]^2[B]. Predict by what factor the rate of reaction will change when the concentration of A is halved, the concentration of B is tripled, and the temperature remains constant. 3 12 0.75 1.25 None of these choices is correct.

Explanation / Answer

5)

Molar mass is the mass of a given substance divided by the amount of that substance, measured in g/mol
given

kf = 1.86 c/m

8 = m(1.86)
m = 4.30mol/kg
4.30mol/kg = xmol/0.2412kg and x = 1.037 mol
53.6g/xg = 1.037mol/1mol and x = 5.17 gm/mol = molar mass

so ans is E

6)
given

the solution is 30% by mass. So we can say that 30% is the same as 30g of solute per 100g of solution. Set this up as a fraction 30g CdSO4/100g solution.

Now you will need to convert the 30g CdSO4 to moles. The information is given to you (MM= 208.46 g/mol) You will need to divide the 30g CdSO4 by 208.46g/mol. You always want to divide by the same unit. This will cancel out the units of grams and give you the units of moles. Next you want to convert the 100g of solution to liters. This will take two steps. 100g solution/1.10 g/mL. This gives you mL. Now convert from mL to L. The ratio there is 1000mL/1L. So you will divide the number you have in mL by 1000 to get L.



30g / 208.46g/mol = 0.143 mol
100 g / 1.10 g/mL / 1000 mL/L = 0.091 L
Now divide 0.143mol/0.091L = 1.58M

SO ANS IS E

7) HERE

osmotic pressure is = MRT
M = moles
R = 0.0821 L*atm/mol*K
T = temperature (in Kelvin)

So since R is in atm and Kelvin, you must convert.
20°C + 273 = 293K
1mmhg = 0.00131atm

there for 28.1mmhg =0.03697 atm

and Moles = Mass/Molar Mass
(assume mass of 0.102g)

now:
0.0369atm = (0.102g/Molar Mass) x .0821 L*atm/mol*K x 293K

Rearrange and solve for Molar Mass:
663 g/mol

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